text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Nov 2019, 18:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Different methods to solve absolute value equations and inequalities

Author Message
TAGS:

### Hide Tags

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

Updated on: 19 Oct 2018, 04:25
8
1
23
Added the PDF of the article at the end of the post!

Different methods to solve absolute value equations and inequalities

Objective of the article
• After going through the article, one will be able to ace all levels of questions on absolute values that GMAT throws at you.

Agenda of the article
• In this article, we will solve 3 examples to learn how to solve absolute value equations easily
• And, we will also solve 3 more examples to understand how we can solve the questions that are related to the absolute value inequalities.
So, let us look at the first example.

Example 1

If x is an integer such that |x - 3| = 7. What is/are the values of x?

Solution

We are given that x is an integer and the value of |x - 3| is 7, and we are asked to find out all the possible values of x.
Now, from the definition of absolute value, we know that:
• The value of |x| = x, if x ≥ 0, and
• The value of |x| = -x, if x < 0.

Note: Many students may wonder, why |x| is written as -x, when absolute value represents the positive value of the number.

• However, if you pay attention to the constraint that “if x is negative”, then |x| is equal to -x.
• And, we all know that negative of a negative number is always positive.
o Hence, |x| is always positive.

Therefore, it is utmost important for you to pay attention to the constraints.

For example: |3| is 3, since 3 is positive.
• And, |-3| is also 3.
o Because -3 is negative. Hence |-3| will be - (-3) which is equal to 3.

Therefore, if we apply the definition of absolute value in the equation |x - 3| = 7, then we’ll get two cases.

Case-1): When x-3 ≥ 0.
• Then, |x - 3| = x - 3, if x – 3 ≥ 0 or x ≥ 3,
o Which also implies that x – 3 = 7 for x ≥ 3.
o Thus, x = 10

Case-2): When x-3 < 0.
• Then, |x - 3| = -(x - 3), if x – 3 < 0 or x <3,
o Which implies, -(x – 3) = 7 for x < 3.
o On multiplying by a negative sign in the above equation, we get:
(x – 3) = -7
o Thus, x = -4
o Hence, the possible values of x are -4 and 10.

Now, if you carefully observe the two cases, then:
• In the first case, we have the equation as x-3=7
• And, in the second case, we got the equation as x-3=-7.
o So, generally, we can say that an equation of the form |x-a|=b will give us two cases:
 x-a=b and x-a =-b
 And, then we can solve it easily to find the value of x.

Let us quickly write down all our learnings from this question so that we can also apply them directly in further question.

Key Takeaways

1. We can see that the process to solve an absolute value equation is very easy. All we need to do is apply the basic concept of absolute values to solve this question.
2. We observed that if we have an equation in the form |x-a|=b then we will get two cases:
a) x-a=b when (x-a) ≥ 0.
b) And, x-a =-b when (x-a) < 0.

With these learnings in mind, let us now increase the difficulty a bit and solve another question.

Example 2

If x is an integer, then find the values of x that satisfy the equation ||x + 9| - 15| = 10?

Solution

As we can see that we are given one modulus inside another modulus:
• That is, |x+9| is inside ||x+9|-15|.
• And, we are asked to find out all the possible values of x such that ||x + 9| - 15| = 10.

So, how do we solve this question?
• Well, we will simply apply the basic properties of absolute values.
1. We will first solve the outermost modulus.
a. And, while solving it, we can substitute the inner modulus with an appropriate variable such as P or Q.
2. Once we have solved the outer modulus, we will get the values for the assumed variable (P or Q) and then we can simplify the inner modulus, by substituting back the modulus.
3. Now, this process will continue even if we had more than 2 moduli in the question.
a. We will first solve the outermost and then we move from outermost modulus to the inner moduli.

So, let’s apply the same process here.
• We will start from the outer modulus.
o Therefore, let us substitute the inner modulus, |x+9|= y.
Hence, we get |y - 15| = 10.

o Now, we will get two cases:
 First, y -15=10 or y=25, when y-15≥ 0 or y ≥ 15.
 Second, y -15=-10 or y=5, when y-15< 0 or y <15.

o Then, we can substitute back y as |x + 9|, in y= 5 and y=25 as shown below.
 |x + 9| = 5………. (i),
 |x + 9| = 25………(ii)
•Thus, we have to solve these two equations and find the value of x.
 Therefore, on solving the first equation, |x + 9| = 5, we get:
• x+9 = 5 or x = -4, when x+9 ≥ 0 or x ≥ -9.
• x+9 = -5 or x = -14, when x+9 < 0 or x < -9.
 Similarly, on solving the second equation, |x + 9| = 25, we will get he two values of x as -34 and 16.

Therefore, the values of x are {-34, -14, -4, 16}

Key Takeaways

1. In this question, we learnt how to solve an absolute value equation involving more than 1 modulus.
a) And, to do that, we always start simplifying the equation from the outer most modulus and then we move to inner moduli.

2. We also learnt that substituting the inner moduli as a variable helps us to solve the question easily.

With these learnings in mind, let us now move to the next question.

Example 3

Find the values of x, if |x - 13| =|2x + 6|, where x is an integer?

Solution

Now, in this question, we have one modulus in the LHS and one modulus in the RHS.
• But, we have only learnt to solve the questions in which we have modulus and constant term on either of the side.
o So, how should we go about to solve this question?
o Well, do not panic. We will try to convert |x - 13| =|2x + 6| in to |x-a|=b form.
 And, how should we do that?
 We can substitute |2x+6|= b and that will give us the equation as:
|x - 13| =b.
• Now, we can simply solve this.
• Therefore, on solving the equation, we will get two cases:
o x-13=b, when x ≥ 13-------- (1)
o And, x-13= -b, when x <13 -------- (2)
• Now, we can substitute back b= |2x+6| in both the equation.

So, let us first substitute in the equation, x-13=b.

Case-1)
• Hence, we get x-13=|2x+6| for x≥ 13
o Let us assume that x-13 is a constant.
o Therefore, we will get two cases:
 First, in which (2x+6) ≥ 0 or x≥-3.
• Second, in which (2x+6) < 0 or x< -3.
 However, don’t forget to consider x≥13.
• So, when x ≥-3 and x ≥ 13, we get 2x+6= x-13
o Now, if x≥-3 and x≥13 then can we directly say that x≥13?
 Yes, we can.
• All the values that are greater or equal to 13 are also greater or equal -3.
 Hence, for x≥ 13, 2x+6= x-13
• Therefore, the value of x is -19.
• Now, tell me, is this possible that for x≥13, we can get x=-19?
o No, right?
o Hence, we can discard this case.

• Now, when x < -3 and x≥13, we get 2x+6= - (x-13)
o But, is it possible that x can simultaneously be less than -3 and greater than 13?
 No, right?
 Hence, we will discard this equation.

Case-2)
Let us now substitute b= |2x+6| in the equation, x-13= -b.

• Hence, x-13 = - |2x+6| or 2x+6= -(x-13), when x <13 and x≥ -3
o Or, 2x+6= -(x-13), for -3 ≤x < 13
 Therefore, x= 7/3 and this also lies in the range.
• But, can we take x=7/3?
o No, we cannot.
o Because, x is given as integer in the question.
o Therefore, we can discard this case.

• Now, when x< 13 and x<-3 then 2x+6= - (-(x-13))
o Since x is less than -3 as well as 13, we can directly say that x< -3.
o Thus, we get 2x+6= (x-13), for x<-3
o Therefore, x= -19 and we can select this value of x as it is less than -3 and also an integer.

Hence, we will get only one value of x and that is -19.

We can also solve this equation by an alternate method.

Alternate method:

Let’s try to solve this question in a different way by applying the definition of |x| = $$\sqrt{x^2}$$

• We are given |x - 13| =|2x + 6|,
• Hence, we can write this as $$\sqrt{(x – 13)^2} = \sqrt{(2x + 6)^2}$$, since |x| = $$\sqrt{x^2}$$
• So, first, let us remove the square root by squaring the equation on both sides.
o Therefore, $$(x – 13)^2 = (2x + 6)^2$$
o And, expanding this, we get: $$x^2 – 26x + 169 = 4x^2+ 24x +36$$
 Which implies that $$3x^2 + 50x – 133 = 0$$
o Or, we can express this quadratic equation as: (x + 19) (3x – 7) = 0
• Therefore, the values of x are -19 and 7/3.
o Since x is given as integer, we can only take x= -19.

Therefore, the value of x is -19.

Key Takeaways

• We solved this question by 2 methods:
o By applying the basic concepts of absolute value
o And, by applying the definition of |x| that it is equal to $$\sqrt{x^2}$$.

• In method-1, we saw that we can substitute any one modulus to be a variable and then solve it to get the equations of the |x-a|=b.
o So, if we are given an equation of the form |x-a|=|x-b|
 Then, we can substitute |x-b|=c and get |x-a|=c
• Which will give us:
o x-a = c for x ≥ a.
o And, x-a= -c for x < a.
• Then we can substitute back |x-b|=c and get the values of x.

• The second method involved a simple application of the definition of |x| = $$\sqrt{x^2}$$.
• You can apply any method to solve this type of questions.
o However, make sure that the value of x you get, always lies within the considered range and also satisfies the constraint given in the question as we saw in case-1.

Now that we have seen different methods of solving absolute value equations by simple application of its definition, let’s see how to solve absolute value inequalities.

Absolute Value Inequalities

Before solving the questions on absolute value inequalities, let’s brush up a few basics of absolute value inequalities.

Consider, |x| > a, where a is a positive number then what is the range of x?
• Well, we can solve this by applying the definition of |x|.
• The inequality |x| > a can be written as:
o x > a, if x ≥ 0 …………. (1)
o -x > a, if x < 0
 Which implies that: x < -a …………. (2)
 Now, on combining (1) and (2), we can say that if |x| is greater than a, then x > a and x < -a.

Now, consider, |x| < a, where a is a positive number, what will be the range of x?
• Applying the definition of |x|,
• The inequality, |x| < a can be written as,
o x < a, if x ≥ 0…………. (1)
o -x < a, if x < 0
 Or, x > -a. …………. (2)
 Now, on combining (1) and (2), we can say that if |x| is less than a, then x > -a or x < a or -a < x < a.

Let’s apply these inferences to solve a few questions.

Example 4

If |x - 8| > 20, what is the range of x?

Solution

This is in the form of |x| > a
Therefore, the range of x is x > a and x < -a.
• Applying the same here, we get that, x – 8 > 20 or x - 8 < -20
o Which implies that x > 28 or x < -12.

Thus, the range of x = (-∞, -12) ⋃ (28, ∞)

Example 5

If ||x - 3| - 8| < 6, what is the range of x?

Solution

We solved a question of double modulus in absolute value equations.
So, we will follow the similar approach.

First, substitute |x - 3| as y, and write the inequality as,
• |y - 8| < 6
• Now, this is in the form of |x| < a
• Therefore, the range of x is -a < x < a.
o Similarly, we can find the range of y if we substitute x by y-8 and a by 6.
 Thus, -6 < y-8< 6
 Adding 8 on all the sides of the inequality, we get: 2< y< 14.

Now, we can substitute back the value of y as |x - 3| and get:
• 2 < |x - 3| < 14
• Here, we need to consider two cases, when x – 3 is positive and when x - 3 is negative
o Considering first case, x – 3 is positive, we have:
 2 < x - 3 < 14
 Simplifying this inequality, we get
• 5 < x < 17 ………….…. (1)
o Now, considering second case, x – 3 is negative, we have:
 2 < - (x – 3) < 14
 So, let us now multiply by -1 on all the sides, we get
• -2 > x – 3 > -14 or -14 < x – 3 < -2

Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by -1?

So, always keep in mind to change the signs when you multiply an inequality by a negative number.

o Simplifying this inequality, we get
 -11 < x < 1 ……………. (2)

Therefore, the range of x is (1) ⋃ (2), which is equal to (-11, 1) ⋃ (5, 17).

Key Takeaways

1. In this question, we learned how to solve an inequality that involves more than 1 modulus.
2. We first substituted the value of inner modulus as y and found the range of y.
o Then, we substituted back the inner modulus at the place of y and considered every possible case that is:
 x – 3 as positive and negative to find the range of x.

With these learnings in the mind, let us discuss our final question of this article.

Example 6

If $$(x + 7)^2$$< 81, what is the range of x?

Solution

It is given that $$(x + 7)^2$$< 81
Now, let us apply the reverse logic of |x| = $$\sqrt{x^2}$$.

Therefore, if you take square root on both sides then we get the inequality as:
• $$\sqrt{(x+7)^2} < \sqrt{81}$$
o Now, we know that $$\sqrt{(x+7)^2} = |x + 7|$$
o Thus, |x + 7| < 9
o And, it is in the form of |y| < a which implies that -a < y < a.
• Therefore, -9 < x + 7 < 9
• By subtracting 7 on all the sides, we get, -16 < x < 2
Therefore, the range of x = (-16, 2)

Key Takeaways from the article

• In this article, we learned that to solve an absolute value equation, we just need to apply the basic definition of absolute value
o That is, |x| = x, if x ≥ 0,
o And |x| = -x, if x < 0.

• We also learnt that if an equation is in the form |x-a|=b then we will get two cases:
o First, when x-a ≥ 0 then we get x-a =b
o Second, when x-a < 0 then we get x-a = -b.

• Then, we learned to follow a methodical approach to solve an equation where we have a modulus inside another modulus.
o First, we solve the outer modulus, by substituting the inner modulus with an appropriate variable, and
o Then, we solve the inner modulus.

• Then, we learned to solve an equation of the |x-a|= |x - b|
o We solved this question by two methods.
 In first method, we substituted one of the modulus as a variable, let say C and then we solved it by using the concept of absolute value.
 In the second method, we used the concept that |x| = $$\sqrt{x^2}$$ and it helped us to find the answer easily.

• We also saw the range of a few absolute inequalities.
o For example: If |x| > a, then we get the range as: x > a and x < -a.
o And, if |x| < a, then we get the range as: -a < x < a

• And, at the end, we saw three questions to understand how we can solve absolute inequality questions.

Now, keeping all these learnings in mind, solve the practice questions.

Want to read more articles like this?

Now, if you liked this article, then I am sure that you want to read some more articles like this.

To read all our article go here:

You will also get a lot of practice questions to learn from.
Happy learning.
Attachments

Absolute value equations and inequalities.pdf [981.63 KiB]

_________________

Originally posted by EgmatQuantExpert on 13 Sep 2018, 02:43.
Last edited by EgmatQuantExpert on 19 Oct 2018, 04:25, edited 3 times in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

13 Sep 2018, 02:46
2
1
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

17 Sep 2018, 05:02
1
Intern
Joined: 29 Aug 2016
Posts: 24
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

17 Sep 2018, 20:03
As always kindly attached the pdf
Intern
Joined: 29 Aug 2016
Posts: 24
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

11 Oct 2018, 06:46
Kindly attach the pdf.

thanks.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

19 Oct 2018, 04:26
Intern
Joined: 29 Aug 2016
Posts: 24
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

29 Oct 2018, 03:06
The PDF doesn't include the Example 6. Kindly revise. Thank you so much.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

31 Oct 2018, 04:40
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

02 Nov 2018, 05:18
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

16 Jan 2019, 02:17
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

18 Jan 2019, 01:53
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

Updated on: 23 May 2019, 21:16

Originally posted by EgmatQuantExpert on 15 May 2019, 02:12.
Last edited by EgmatQuantExpert on 23 May 2019, 21:16, edited 1 time in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

20 May 2019, 04:39
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

26 Jun 2019, 02:56
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

26 Jun 2019, 03:46
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

31 Jul 2019, 05:19
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3142
Re: Different methods to solve absolute value equations and inequalities  [#permalink]

### Show Tags

05 Aug 2019, 22:03
Re: Different methods to solve absolute value equations and inequalities   [#permalink] 05 Aug 2019, 22:03
Display posts from previous: Sort by


text:
# Product of normal random variables - bivariate normal?

Im wrong about something here, but Im not sure what.

As far as I know the product of two normal distributed variables is not normal distributed. However, if the joint distribution of Y and X is bivariate normal then Y given X is normal distributed as well as the marginal distribution of X. Furthermore, the joint distribution of Y and X is the product of the condtional distribution of Y given X and the marginal distribution of X which are both normal. If we define a new variable Z to be distributed as the condional distribution of Y given X, then the product of Z and X (which are both normal distributed) is bivariate normal.

Am I on the right track or did I miss understand something somewhere?

You're talking about two different things:

$$V=XY$$

vs.

$$f_{XY}(x,y)$$

The product of two variables cannot be bivariate, since a product is a number.

format: plain


text:
• # question_answer If all the sides of a triangle be increased by 200 per cent, what is the corresponding increase in its area? A) 300%                           B) 400% C) 600%                           D) 800%

Let sides are a, b, c. So, new sides are 3a, 3b, 3c. Original area $=2l+2b$ So, new area $=2(l+b)$ Increased area 9 K - K = 8K % = 800%.


text:
# How do you solve sqrt(x-3)-sqrtx=3?

Jul 28, 2016

There are no solutions.

#### Explanation:

The equation given can be reformed to give:
$\sqrt{x - 3} = \sqrt{x} + 3$

This is equivalent to asking where two functions intersect. The functions in this case are:
$y = \sqrt{x - 3}$
$y = \sqrt{x} + 3$

Merely observing the graph of the functions makes it clear that the two will never intersect:
graph{(y-sqrt(x-3))(y-sqrt(x)+3)=0 [-10.97, 46.77, -9.94, 18.93]}

You may note that the functions appear to head toward one another at $x = 0$. At this point the functions both become imaginary valued. If the graph were continued in complex space, they still would not intersect.


text:
# International Journal of Philippine Science and Technology ( formerly Philippine Science Letters )

Posted on in the International Journal of Philippine Science and Technology 2015 Volume 8 Issue Number 2 Page 3135

*Corresponding Author
Submitted: May 18, 2014
Revised: May 29, 2015
Accepted: June 16, 2015
Published: July 28, 2015

ARTICLE

### by Severino G. Salmo III*1, and Dranreb Earl O. Juanico2,3

##### 1Department of Environmental Science, Ateneo de Manila University, Loyola Heights, 1108, Quezon City, Philippines2Technological Institute of the Philippines, Quiapo, 1001, Manila, Philippines3The Oscar M. Lopez Center, Ortigas, 1605, Pasig City, Philippines

Abstract—We developed a spatially explicit individual-based model of forest development trajectories of monospecific Rhizophora mucronata plantations. The model incorporates stochastic initial seedling spacing, propagule dispersal, recruitment, and mortality. We simulated and compared the growth, development and accumulation of carbon stocks of restored mangroves between optimal and sub-optimal settings. Salinity is considered as a stressor, while flooding effects are described as an inundation stress factor. In the optimal setting, the simulated mangrove population accumulated large aboveground carbon stocks (of around 140 T/ha) after 100 years. Under sub-optimal conditions, we observed delayed maturity by at least 10 years near the salinity threshold and the carbon stock decreased through time towards much lower values (25 T/ha). More importantly, the continuous presence of stressors may lead to forest population collapse (at 50 yrs post-planting) probably as a result of the accumulated effects of physiological stresses. Thus, restored mangrove populations that are located in highly saline and frequently inundated sites, may eventually collapse even though they may appear to be healthy in the early stages of forest development. Our results imply that current and future restoration practices should carefully consider site selection in order to ensure viable long-term forest development and to have an optimum contribution to carbon sequestration.

Keywords—mangroves, individual-based model, growth, forest development, restoration trajectory, salinity, inundation, carbon sequestration

##### INTRODUCTION

Mangrove restoration programs have long been implemented in the Philippines (Walters 2003) receiving funding from the local government, national government, non-government organizations and international funding agencies. Mangrove restoration has recently been considered as a climate change adaptation and mitigation strategy because of its potential role in the sequestration of atmospheric CO2 (Donato et al. 2011). The effectiveness of a mangrove restoration program, however, depends on the growth and survival of the planted mangroves and, subsequently, the long-term stability of the forest.

In the Philippines, most mangrove plantations are located in sub-optimal settings that are highly saline and frequently inundated (Samson and Rollon 2008). The coastal fringe is considered sub-optimal for mangrove planting. Most restoration programs commonly use Rhizophora spp. due to their relative ease of sourcing and transplanting propagules as compared to the more appropriate Sonneratia or Avicennia spp. But Rhizophora mucronata, in particular, is known to be stenohaline and thus disadvantaged in areas where tidal frequencies (hence, salinity fluctuations) are high. Waves further make the coastal fringes less suitable for R. mucronata as wave action will erode the substrate and reduce the stability of the planted seedlings. Most mangrove planting programs are implemented in numerous sites deemed biophysically unsuitable for R. mucronata. Primavera and Esteban (2008) cite policy-related reasons for the persistence of such restoration malpractice. Most of the planted seedlings have stunted growth and poor survival. However, monitoring the growth and survival as well as the success or failure of these mangrove restoration programs is largely not undertaken. If any monitoring is conducted, the findings are rarely reported.

Due to the long periods of time needed for the growth and development of mangrove forests, assessment of the success of mangrove restoration programs is challenging. Mathematical models offer a way to circumvent this challenge. However, available mangrove forest models such as KiWi (Berger and Hildenbrandt, 2000) and FORMAN (Chen and Twilley 1998) have built-in assumptions regarding environmental conditions that are more applicable for sub-tropical mangrove systems (Berger et al. 2008). In their original configuration, these existing models are not ideal for mangrove systems in the Philippines, which thrive under remarkably different environmental conditions. On the other hand, the MANGRO model (Doyle and Girod, 1997) is more applicable to the Philippine setting because it considers the effect of tidal flooding on tree growth. However, it does not provide a mechanism for seedling or sapling dispersal (Berger et al. 2008).

Here, we used model simulations to compare the long-term trajectories of forest development and carbon stock accumulation between settings that mimic typical reforested sites in the Philippines and those considered biophysically optimal for R. mucronata. The model we developed for this study combines the strengths of KiWi in describing resource competition and salinity sensing, and of MANGRO in considering tidal inundation as a stressor that attenuates the growth rate and survival of individual mangroves. In addition, our model incorporates stochastic recruitment, dispersal, and mortality, which are not considered simultaneously in any of the said models. We do not account for wave exposure in this study by assuming that extreme wave action does not occur frequently. The biophysical parameters of planted R. mucronata required to specify the model inputs are deduced by fitting simulation results with actual growth monitoring data obtained from our recent field observations (Salmo III et al. 2013). We tracked the development trajectories of virtual mangrove forests over a period equivalent to 100 years, although the model could be simulated over longer periods of time.

##### Model development

The model is a spatially explicit, individual-based model founded upon a description of the rate of change of the diameter-at-breast-height, dbh. The dbh is used as a gauge for the growth of individual mangroves. Allometric relations can estimate the growth rate of other aspects of mangrove structure (i. e., crown diameter) and biomass by extrapolating dbh (Botkin 1972). In the model, growth stressors are represented by a salinity stress factor and inundation stress factor where the value 1 indicates the absence of the stressor (hence, optimal condition). Competition between individual mangroves is computed based on the Field of Neighborhood (FON) approach and the Zone of Influence (ZOI) concept in the KiWi model (Berger and Hildenbrandt 2000). In the presence of stressors and resource competition, the growth of an individual mangrove is given in Equation 1 (for convenience, dbh is renamed as D), as:

$\frac{dD}{dt}={\left&space;(\frac{\Omega}{2+\alpha}\right&space;)D}^{\beta-\alpha-1}\left&space;[&space;1-\frac{1}{\gamma}{\left(\frac{D}{D_{max}}&space;\right&space;)}^{1+\alpha}&space;\right&space;]*C*S*E$      (1)

where α = 0.95 and β = 2 are allometric constants for height and tree crown, respectively, as a power function of D. The constant Ω, is a conversion factor with a value of 0.25 to harmonize the left and right sides of Equation 1. The values for α and Ω were derived from the growth and biomass of planted mangroves as reported in Salmo III et al. (2013) while β was derived from the allometric relation of D and leaf area index (LAI; Botkin 1972). Thus, the simulation of a given time scale (i.e. day, month, year) can be changed without necessarily losing its generality. Meanwhile, γ and Dmax are set based on the maximum possible height Hmax = 130 m (Koch et al. 2004). Because an individual mangrove is confronted by resource competition in a forest, the growth rate is slower on average within a forest or within patches of trees than in isolation. Lastly, the factor C represents the attenuation to the growth rate of an individual mangrove caused by competition with other mangroves. Competition is further explained in a later section.

The stress factors S and E due to salinity and inundation, respectively, are valued between 0 and 1. Coastal fringe conditions are represented by 0, whereas mid-intertidal conditions by 1. Based on Equation 1, a value close to zero would decrease growth with time such that the individual mangrove grows at a slower pace. In particular, E attenuates the average growth rate because of stress caused by submergence (partially or totally) under waterlogged condition. The diurnal cycles of the tides also cause fluctuations in salinity levels. Although tidal flooding and salinity appear correlated, we assume in our model that these factors are separable. While this assumption adds tractability, it is nevertheless a limitation of the model. In our model, we set the mangroves to grow optimally if salinity is ≤ 25 ppt and will be stressed if salinity will be 27 ppt (S = 0.38), 29 ppt (S = 0.27) and 40 ppt (S = 0.02). The associated values of S are determined from salinity x by a sigmoidal response function: S = S(x) = {1+exp[0.25(x – 25)]}–1. Similarly, growth of mangroves will be optimal if they are not inundated (E = 1; representing mean tidal level) but will be constrained if they are inundated for 4-6 hours (E = 0.90), 6-8 hours (E = 0.75) and 8-12 hours (E = 0.50) per 24-hr cycle (Table 1).

Table 1. Growth factors due to R. mucronata for salinity (regulator) and inundation (hydroperiod) in optimal and sub-optimal settings.

We explored the long-term growth, development, and carbon stock accumulation of restored R. mucronata forests by comparing different salinity and inundation settings. Each setting is mathematically represented as an ordered pair (S, E) of the two factors listed in Table 1. The software implementation of the model was made using the C++ programming language on a Unix platform in a 1.7 GHz Intel Core i5 machine. A detailed description of a related model using the “Overview-Design-Detail” (ODD) framework is provided in Juanico and Salmo III (2014).

##### Initialization

A 40m × 40m (= 0.16 ha) mangrove patch is initialized. The simulation moves forward with an annual time step (∆t = 1 year). In the beginning, there are 100 seedlings of R. mucronata with initial dbh of 0.5 cm that are scattered somewhat randomly at 1.0 – 1.5 m distance apart. Thus, the distance of one seedling to another is about 1.0 – 1.5 m, which approximates the typical planting density for most mangrove restoration sites in the Philippines (Samson and Rollon, 2008).

The simulated mangrove forest was assumed to have uniform, unchanging salinity level. We also assumed that the tidal exposure of the forest leads to a uniform, unchanging inundation stress. Here, we consider inundation stress as a time-averaged effect of tidal exposure over a long period of time. Note that inundation stress does not denote tidal cycle. Since the mangrove plantation is monospecific, then all parameters listed in Table 1 were fixed at the beginning of a simulation. The simulations were run for five times for each S and E setting.

##### Growth

Equation 1 was numerically solved for each mangrove in the patch. Using ∆t = 1, the solution was implemented using Euler’s forward scheme. Consequently, the dbh of an individual mangrove was increased by an amount approximately equal to the right-hand side of Equation 1 at time t. The life stage of the mangrove was determined by means of the following standard classification based on dbh: seedling, 0.50 ≤ dbh < 2.50 cm; sapling, 2.50 ≤ dbh < 5.0 cm; and tree, dbh ≥ 5.0 cm. By implementing a positivity constraint on Equation 1, the value of dbh was only expected to increase as time progresses.

##### Recruitment

Based on our field data (Salmo III et al. 2013), an annual average of five seedlings from each mature R. mucronata tree was recruited. Established seedlings have initial dbh of 0.50 cm, and were scattered randomly within a 5 m radius around the parent tree. The recruitment by each mature mangrove was also taken as a stochastic process with a constant probability per unit time. Therefore, the total number of recruits at any given time was a random variable. This randomness reasonably accounted for the probabilistic survival of recruits due to many other confounding factors, e.g., predation, disease, or wave-current action, which were not explicitly accounted for in the present model.

##### Mortality

The loss of individuals due to mortality was described as a stochastic process with a probability of death that approximated the total mortality rate of the entire population. Due to confounding factors, such as predation or pest infestation, each individual mangrove has a constant chance of dying at any given point in time. The rate of mortality becomes higher as tree density increased due to more intense resource competition or as the available growing spaces are occupied (cf. Oliver and Larsson 1996).

Mangrove mortality rates are expected to conform with Type III survivorship curves, as any plant in general (Schaal and Leverich 1982). The life stage (e.g., seedling, sapling or tree) determines the mortality rate. Seedlings have the highest mortality rates, whereas trees have the lowest. Dying in the population is treated as a Poisson process. Any mangrove has a chance of dying at any given point in time. The probability of dying within a certain time period of a mangrove at a given life stage is related to its mortality rate. Because progression through life stages by growth is constrained by resource competition, the probability of dying is indirectly related to population density.

##### Competition

We modeled tree-to-tree competition following the approach in the KiWi model. One of the strengths of the KiWi model is its generic approach in describing resource competition in a forest. We followed closely the FON approach for calculating the intensity of competition based on the dbh size and the position of an individual mangrove in the forest (Berger and Hildenbrandt 2000). In this method, an individual mangrove is affected by competition with its neighbors. The effect on the individual is a slowdown to its growth. Neighbors at a distance contribute to this effect through an extended root system below the ground and a tree crown aboveground. The aggregate effect on an individual from all other mangroves is best described mathematically using field variables. If the aggregate field on a mangrove at position (x,y) is F, then the attenuation factor it contributes toward the slowdown of the growth rate is C(x,y) given below:

$C(x,y)=1-2F(x,y)$      (2)

##### Carbon stock

Biomass evidently increases as the mangrove grows. Because the model tracks growth by observing the dbh, we used an allometric equation (cf. Komiyama et al. 2008) to convert a mangrove’s dbh to above-ground biomass (AGB, in kg per tree):

$AGB=0.235\times&space;dbh$      (3)

##### Dispersal

A propagule from a tree may establish and grow anywhere within an annular region surrounding the tree. This annular region has an inner radius equal to the parent tree's crown radius. On the other hand, its outer radius is equal to the product between time and dispersal rate. The dispersal rate is species-specific. For R. mucronata, we assume that this rate is about 26.67 cm per day deduced from measurements by Sousa et al. (2007). Stochasticity was incorporated by assigning a random position with respect to the vertical axis of the parent tree. The stochastic dispersal addresses the uncertainty in tracking the exact site where a seedling establishes.

##### Spatial distribution

The snapshots of mangrove forest growth and development under the optimal setting, (S, E) = (1, 1) for t at 5, 25, 50, and 100 years are shown in Figure 1. The perimeter enclosing all mangroves is called the convex hull where all mangrove individuals (seedlings, saplings and mature trees) per simulation are confined. The values for tree density, aboveground biomass, and carbon stocks were computed based on the set area of the mangrove patch. The use of the convex hull, rather than the entire patch, shows a more accurate description of the actual spatial distribution of mangroves. The density of mature trees increased with time. As time progressed, there was an apparent increase in the number of mature trees accompanied by the decrease in available growth spaces as a result of tree-to-tree competition. The fact that the tree density is increasing at the optimal setting (1,1) implied that AGB also increased as time progressed.

Figure 1. Development of restored mangrove forest over 100 years for the optimal setting (S = 1, E = 1). The zero (0) point in the Y-axis denotes the edge of the shoreline. Each graph represents a snapshot of mangrove distribution 5, 25, 50, and 100 years post-planting. Each dot represents the top view of the crown (computed from dbh) of an individual mangrove. The indicated perimeter is the convex hull, which encloses the area occupied by all mangrove individuals.

##### Spatial distribution

The snapshots of mangrove forest growth and development under the optimal setting, (S, E) = (1, 1) for t at 5, 25, 50, and 100 years are shown in Figure 1. The perimeter enclosing all mangroves is called the convex hull where all mangrove individuals (seedlings, saplings and mature trees) per simulation are confined. The values for tree density, aboveground biomass, and carbon stocks were computed based on the set area of the mangrove patch. The use of the convex hull, rather than the entire patch, shows a more accurate description of the actual spatial distribution of mangroves. The density of mature trees increased with time. As time progressed, there was an apparent increase in the number of mature trees accompanied by the decrease in available growth spaces as a result of tree-to-tree competition. The fact that the tree density is increasing at the optimal setting (1,1) implied that AGB also increased as time progressed.

##### Long-term carbon stock accumulation

Based on the different possible ordered pairs (S, E) of values listed in Table 1, the time-averaged levels of aboveground carbon stock at 100 years increased with time [n = five replicates of each (S, E)]. Figure 2 depicts the long-term mean carbon stock of the forest under different settings provided in Table 1 (S = {0.02, 0.05, 0.10, 0.27, 0.38, 1.00} and E = {0.50, 0.75, 0.90 and 1.00}. The average aboveground carbon stock at 100 years for the optimal setting (1,1) is about 140 tC/ha. Compared to the sub-optimal settings, the optimal result was found to be considerably higher by at least 60 tC/ha. Generally, for any value of S considered, the trend for carbon stock with respect to E was positive except for S = 0.38. On the other hand, the long-term carbon stock significantly declined when S was less than 0.1, to levels at least an order of magnitude lower than the optimal value (Figure 2).

Figure 2. Long-term carbon stock accumulated after 100 years for different (S, E) settings. Each data point represents the average of five simulations for a given setting. The error bars indicate the standard deviation from the mean. The deviation from the expected monotonic trend for (0.27, 1) and (0.38, 1) can be explained in part by the salinity response S(x). The sigmoidal graph of S(x) puts the salinity levels associated with S = 0.27 and S = 0.38 at the transition part. The transition is associated with more statistical uncertainty.

There was a pronounced difference in the carbon stock trajectories between the optimal setting and the most extreme sub-optimal setting. Simulations showed a continuous accumulation of carbon stock due to increasing tree density for up to 100 years since initial planting for the optimal setting (Figure 3). In contrast, in the most extreme sub-optimal setting (S = 0.02; E = 0.05), seedling density decreased rapidly within the first 10 years, then the remnant population (composed of surviving saplings that eventually matured as trees) finally collapsed around 50 years (Figure 4B). Consequently, the carbon stock in living mangroves was non-existent after 50 years (Figure 4A). For sub-optimal settings, S and E are smaller than unity, so that plants would linger in the less mature state for a longer time (see Equation 1). Since immature trees are more susceptible to mortality than mature trees, then other instantiations of the sub-optimal settings are likely to lead to decreases in populations and carbon stock over time.

Figure 3. Trajectory of (A) carbon stock, and (B) density of trees, saplings and seedlings over 100 years in a single realization of the model for the optimal setting (S = 1, E = 1).

Figure 4. Trajectory of (A) carbon stock, and (B) density of trees, saplings and seedlings over 100 years in a single realization of the model for the extreme sub-optimal setting (0.02, 0.50).

##### Effect of salinity stress

Simulations suggested that the density of mature trees (> 0.5 cm dbh) under different values of S (with a fixed value E = 1) will increase through time for S = 0.27, 0.38, and 1.0 (Figure 5). The forest growth rate is indicated by the slope of the trajectory, which is most positive (i.e., highest rate of increase in tree density) for the optimal case S = 1. On the other hand, below an unidentified threshold (ca. 0.02 < S < 0.27), the trajectory trends downward or the slope is negative. For S = 0.02, the development of mature trees occurred at a later time than for the other less stressful S values. The delay could be explained by the slow growth, and thus slow transition, of individual mangroves (from seedling to sapling then from sapling to tree) in a sub-optimal salinity setting even in the absence of inundation stress.

Figure 5. Trajectory of changes in tree density with time for the different values of the salinity stress factor S at the optimal value of the inundation stress factor, E = 1. Each line is a single realization of the (S, E) setting.

##### Effect of inundation stress

The density of mature trees (with a fixed value S = 1) had an upward trajectory through time regardless of E values considered (Figure 6). The slope was highest for the optimal case (E = 1), indicating fastest forest growth rate. The highest tree density 100 years post planting was achieved at the optimal value. Unexpectedly, tree density after 100 years was higher for E = 0.5 as compared to E = 0.75. This may have been an artifact of the simulations due to stochasticity. The fact that tree density had a consistent upward trend regardless of E values considered strongly suggests that R. mucronata is less sensitive to wide variations to inundation stress compared to salinity stress.

Figure 6. Trajectory of changes in tree density with time for the different values of the inundation stress factor E at the optimal value of the salinity stress factor, S = 1. Each line is a single realization of the (S, E) setting.

##### Individual-based model for restored monospecific mangroves

Our model provides a platform that analyzes the restoration trajectory of planted monospecific mangroves, in this case of the species Rhizophora mucronata. We adapted the Field of Neighborhood approach and the concept of Zone of Influence from the KiWi model (Berger and Hildenbrandt 2000). However, we incorporated several improvements based on actual R. mucronata plantation data in the Philippines (Salmo III et al. 2013) and adjusted other model assumptions to approximate Philippine environmental conditions. Hence, our model can be used to project the possible development or demise of typical mangrove restoration programs in the Philippines.

We evaluated the effects of different levels of salinity and inundation in the model. These parameters are the main stressors that can strongly constrain the growth and survival of planted mangroves (see Samson and Rollon 2008 for example). The growth of mangroves, particularly of the species R. mucronata, is physiologically sensitive to high salinity (>25 ppt) and prolonged periods of inundation (>4 hours submergence per 24-hour cycle). The salinity and inundation settings that we used in the model were derived from long-term field observations (10 years or more) at various mangrove restoration sites across the Philippines (Salmo III et al. 2013). In our simulations, we evaluated the effects of salinity levels that were not higher than 40 ppt. Note, however, that most replanted sites can have extreme salinity levels of up to 60 ppt, which will have stronger adverse effects on mangrove growth and survival.

We also modified the mechanism of propagule dispersal, mortality and stochasticity from the KiWi model. The propagules are transported and able to successfully establish only within a 5 m radius from the parent tree. This limited range of dispersal is consistent with observations in the field. This is probably due to the limited buoyancy of the propagules as well as the fact that propagules can often be stranded in the dense network of prop roots of R. mucronata. Mortality in the model is probabilistic. The probability of dying is negatively associated with age and positively associated with the rate of increase in the diameter of an individual tree, hence mortality rate is higher for seedlings and lower for mature trees. In terms of stochasticity, we did not define the time to which an individual mangrove can transition from seedling to sapling or from sapling to tree, but instead let the change in dbh determine its life stage.

Our modifications of the KiWi model are consistent with our field observations. However, we acknowledge some possible limitations of our model. Factors such as spatial gradients in salinity and inundation levels as well as the strength and direction of water currents in the entire patch relative to distance from the shoreline will obviously have an effect on the growth and survival of each individual mangrove. These limitations are currently being addressed in an ongoing complementary study. Nonetheless, we believe that the present model was able to capture the general pattern of growth of individual mangrove and critical parameters in the dynamics of mangrove populations. This would then enable the simulation and assessment of long-term restoration trajectories of planted mangroves.

##### Optimal vs. sub-optimal conditions

As expected, the restoration trajectories (in terms of changes in tree density, aboveground biomass and carbon stock with time) of mangroves varied widely between optimal and sub-optimal settings (Figures 2, 4, 5 and 6). Under optimal conditions, the mature trees (> 5 cm dbh) appeared to be sporadically distributed at the 5th and 25th yr but tend to be more congregated on the 50th and 100th yr (Figure 1). Such a distribution pattern was probably due to the limited range of propagule dispersal (only 5 m from the parent tree) that we imposed for R. mucronata in the model. These results are also consistent with our field observations. However, they contrast with the results of simulations by Berger and Hildenbrandt (2000) for R. mangle, a neotropical mangrove species. In their study, the trees were almost uniformly distributed until the available growing spaces (cf. Oliver and Larsson 1996) were occupied. Such contrasting results between the two studies can be attributed to the effect of inundation (which was not explicitly accounted for by Berger and Hildenbrandt 2000) as well as the combined effects of salinity and inundation. The presence of these stressors will most likely result in better establishment and growth of individual trees in areas closer to the shoreline.

The trajectories suggest that the density of mature trees rapidly increases within a very short time (<10 years) to about 150-400 trees/ha in all scenarios, except when S = 0.2, E = 1 (Figures 5 and 6). Furthermore, the increase was apparently the steepest for the optimal setting, S = 1, E = 1, between the 10th and 70th years. Compared to the baseline, the increase in the trend for sub-optimal settings is relatively flatter in the said period.

The simulated aboveground biomass and carbon stocks here are remarkably lower compared to published reports (see Donato et al. 2012 for example). However, the low carbon stock values are similar to those reported for Philippine mangrove plantations (55 – 60 tC/ha; Salmo III et al. 2013). Even in the optimal settings (S = 1, E = 1), the biomass and carbon stocks are also low (between 71 and 209 tC/ha at 100 years in the simulations) which can probably be accounted for by the low diversity of the restored system (see also Isbell et al. 2013). Also, the model implies that to have optimum growth, Rhizophora seedlings must be planted above the mean tidal level.

A striking result of our simulations is the sudden collapse of the mangrove population (and subsequently carbon stock) at 50 years post planting particularly under the most extreme sub-optimal conditions (Figure 4). Such a population collapse, while still largely undocumented for current mangrove restoration sites in the Philippines, makes sense ecologically because of the persistent presence of physiological stressors that could eventually lead to the demise of the planted mangroves. A monospecific plantation is known to be less resilient against stressors as compared to a more diverse mangrove forest system (Gunderson 2000). Thus, while we acknowledge that mangrove restoration programs can contribute to carbon sequestration, the amount and rates of carbon sequestered by typical mangrove restoration efforts are likely to be very low and there is a possibility that the planted mangroves may perish in the long run.

##### Management implications

Mangrove restoration has long been practiced in the Philippines (see Talaue-McManus et al. 1999; Walters 2003; Salmo et al. 2007; Primavera and Esteban 2008; among others). However, very few successful cases have been reported. Many programs have failed basically because of inappropriate species-substrate matching and planting in unsuitable locations (Samson and Rollon 2008). These monospecific restoration programs have long been criticized due to the stunted growth and poor survival of mangroves and their potential effect of reducing mangrove species diversity (Primavera and Esteban 2008).

The use of mangrove restoration programs to mitigate the impacts of global warming seems like a viable strategy and provides additional incentives for mangrove managers. Planted mangroves can of course contribute to the sequestration of atmospheric CO2. However, restored mangroves will be effective in carbon sequestration over the long term only if the mangrove population will have steady growth and maintain its stability. Unfortunately, this is not the case yet for most mangrove plantations in the Philippines.

In our simulation results, we demonstrated that mangroves (Rhizophora sp.) planted in sub-optimal conditions would have much lower levels (up to 100%) of aboveground biomass and carbon stock 100 years post-planting compared to those planted in optimal conditions due to mortality. In addition, the population collapse that may occur around 50 years post-planting implies the need for frequent replanting at the same site in order to maintain the population. Thus, although the planted mangroves may initially appear to be healthy (for example even at 20 years post-planting), there is a possibility that the population will collapse in the long run because of the persistence of stressors and the cumulative effects of physiological stresses. Therefore, the accumulated carbon stocks in the restored mangrove systems planted in sub-optimal conditions may be in danger of being lost over time. Hence, we emphasize the importance of the appropriateness of the planting site in terms of the levels of salinity and inundation stresses. There are other factors that have not been considered explicitly, such as wave stress. However, in the model we presented here, we focused on evaluating the impacts of extreme salinity and inundation stress levels in the backdrop of tree competition and seed dispersal. Despite its limitations, the model offered some important insights about long-term forest growth and carbon sequestration. We considered the presence of stressor gradients in a separate study (see Juanico and Salmo III 2014).

##### ACKNOWLEDGEMENTS

This research was funded by the Commission on Higher Education (CHED) – Philippine Higher Education Research Network (PHERNet), and was presented at the PAMS 12 conference in Tacloban City in October 2013.

None

##### CONTRIBUTIONS OF INDIVIDUAL AUTHORS

Dr. Salmo conceptualized and implemented the study. Dr. Salmo wrote the introduction and discussion sections. Dr. Juanico developed the model and analyzed the results of simulations. Dr. Juanico also wrote the methods and results sections.

##### REFERENCES
• Berger U, Rivera-Monroy V, Doyle TW, Dahdouh-Guebas F, Duke NC, Fontalvo-Herazo M, Hildenbrandt H, Koedam N, Mehlig U, Piou C, Twilley RR. Advances and limitations of individual-based models to analyze and predict dynamics of mangrove forests: a review. Aquat Bot 2008: 26-274.
• Berger U, Hildenbrandt H. A new approach to spatially explicitly modelling of forest dynamics: spacing, ageing and neighbourhood competition of mangrove trees. Ecol Model 2000; 132:287-302.
• Botkin D, Janak JF, Wallis JR. Some ecological consequences of a computer model of forest growth. J Ecol 1972: 849-872.
• Chen R, Twilley RR. A gap dynamic model of mangrove forest development along gradients of soil salinity and nutrient resources. J Ecol 1998; 86(1): 37-51.
• Donato DC, Kauffman JB, Murdiyarso D, Kurnianto S, Stidham M, Kanninen M. Mangroves among the most carbon-rich forests in the tropics. Nat Geosci 2011; 4:293-297.
• Doyle, TW, Girod, GF. The frequency and intensity of Atlantic hurricanes and their influence on the structure of south Florida mangrove communities. In Hurricanes 1997: 109-120. Springer Berlin Heidelberg.
• Gunderson LH. Ecological resilience: in theory and practice. Annu Rev Ecol Syst 2000; 31:425-439.
• Isbell F, Reich PB, Tilman D, Hobbie SE, Polasky S, Binder S. Nutrient enrichment, biodiversity loss, and consequent declines in ecosystem productivity. Proc Natl Acad Sci USA 2013; 29:11911-11916.
• Juanico DE, Salmo III SG. Simulated effects of site salinity and inundation on long-term growth trajectory and carbon sequestration in monospecific Rhizophora mucronata plantation in the Philippines. ArXiV Preprint 2014; arXiv:1405.6944
• Koch GW, Sillett SC, Jennings GM, Davis SD. The limits to tree height. Nature 2004; 428: 851-854.
• Komiyama A, Ong JE, Poungparn S. Allometry, biomass and productivity of mangrove forests: a review. Aquat Bot 2008; 89: 128-137.
• Masera OR, Garza-Caligaris JF, Kanninen M, Karjalainen T, Liski J, Nabuurs G, Pussinen A, de Jong BHJ, Mohren GMJ. Modeling carbon sequestration in afforestation, agroforestry and forest management projects: the CO2FIX V.2 approach. Ecol Model 2003; 164:177-199.
• Oliver CD, Larsson BC. Forest stand dynamics. Wiley Inc., New York, USA. 1996; 520 pp.
• Primavera JH, Esteban JMA. A review of mangrove rehabilitation in the Philippines: successes, failures and future prospects. Wetl Ecol Manag 2008; 16(5): 345-358.
• Schaal BA, Leverich WJ. Survivorship patterns in an annual plant community. Oecologia 1982; 54: 149–151.
• Salmo III SG, Lovelock CE, Duke NC. Vegetation and soil characteristics as indicators of restoration trajectories in restored mangroves. Hydrobiologia 2013; 720:1-18.
• Salmo III SG, Torio DD, Esteban JMA. Evaluation of rehabilitation strategies and management schemes for the improvement of mangrove management programs in Lingayen Gulf. Sci Diliman 2007; 19(1): 24-34.
• Samson MS, Rollon RN. Growth performance of planted mangroves in the Philippines: revisiting forest management strategies. Ambio 2008; 37:234-240.
• Sousa WP, Kennedy PG, Mitchell BJ, Ordonez, BM. Supply-side ecology in mangroves: do propagule dispersal and seedling establishment explain forest structure? Ecol Monog 2007; 77: 53-76.
• Talaue-McManus L, Yambao AC, Salmo III SG, Aliño PM. Bolinao, Northern Philippines: Participatory planning for coastal development. In: Buckles D, ed. Cultivating Peace: Conflict and Collaboration in Natural Resource Management. International Development Research Center/World Bank: Ottawa, 2000: 151-161.
• Walters BB. People and mangroves in the Philippines: fifty years of coastal environmental change. Environ Conserv 2003; 30(2): 293-303.

### Announcements

STRIDE Grants The next call for Grants will commence on February 27, 2015. To know more about the Grants program of STRIDE, you may visit our website at stride.org.ph.


text:
1,070 Answered Questions for the topic Math Equations

Math Equations

04/12/13

7.8 x 10^6
Math Equations

04/12/13

Math Equations

04/12/13

Math Equations

04/12/13

Math Equations

04/09/13

#### Simplify the expression

(-2p)^4 What is the simplified expression ?
Math Equations

04/08/13

#### What is the equation of the line that passes through the points (2, 5) and (1, 2)?

I really never got this type of problem, I never really got graphs and slopes
Math Equations

04/07/13

#### 10m^4* (2m^5)^6

How do you solve it?
Math Equations

04/07/13

If $80 is charged for a design fee, and the monthly studio rent is 1,600; write an equation for the profit, P, in terms of x. Then how mch is the profit when 50 award designs are sold in a month?... more Math Equations 04/05/13 #### -v/7 + 8 = 19 ( The slant ( / ) is meant for the fraction not division) Math Equations 04/05/13 #### Use the elimination method to solve the system of equations. a+6b=13 a+4b=7 Select the correct choice (a) There is one solution, it is what (Simplify your answer type a ordered pair must be a ordered pair) (b) There are infinitely many solutions (c)... more Math Equations 04/05/13 #### An investor invested a total of$2,600 in two mutual funds.

One fund earned a 7% profit while the other a 3% profit. If the investor's total profit was \$138, how much was invested in each mutual fund?         The amount that was invested in the mutual... more
Math Equations

04/05/13

#### 3x^2 +2x -16 =0

solve the quadriatic equation using the quaratic formula, no matter how it is done it comes up wrong and nothing is working
Math Equations

04/04/13

#### Solve the system of equations

(a) symbolically, (b) graphically, (c) and numerically 3x+y=6 x-y=2 There are different methods for solving a system linear equations symbolically. Here we will use elimination method. What is... more

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.


text:
Categories

# pythagoras circle problems

The problems range in difficulty: Qs 1, 2, 6, 7 are simple, Qs 4, 5 are more complex Qs 3 and 8 are challenging. A selection of problem cards with real life Pythagoras problems. A parallelogram is formed by joining together four equilateral triangles. If the midpoints of the sides of a right angled triangle are joined, what is the perimeter of this new triangle? This is a graphic, simple and memorable way to remember the difference from a chord or a tangent or a segments and sectors! Let's take the width as (x + 4) cm. University of Cambridge. Copyright © 1997 - 2021. 3464 0 obj <> endobj Pythagoras and Circle Area If two of the sides of a right-angled triangle are 5cm and 6cm long, how many possibilities are there for the length of the third side? Pythagorean theorem - math word problems Number of problems found: 723. Solve two challenging problems that apply properties of tangents to find the radius of a circle with a tangent. Word Cloud of Pythagorean Theorem: Einstein and Pythagoras theorem proof . Find the length of its diagonal. Further the the radius has been stated, but not marked on. A square has area 72 cm$^2$. What is the height of the piece that is still standing? Can you find the length and width of the screen of this smartphone in inches? Problem 1: A 35-foot ladder is leaning against the side of a building and is positioned such that the base of the ladder is 21 feet from the base of the building. H���Mo�@���+�Rٝ��)��JU�N�RVc�5�����R��˰�<3��������F��}Q�l�k� m߷'��}�˱>��$���V!��̽�+����k����.�90�9���]�A(/ ���asFW�=�_�jbs.��*X��x�Fzr��-)�X� �@}�F&����1᪗޿�ZA�*(_AND��9�3�gR�,�ȗ�:�V�B�Qր;� B��'�1I�1��W�N%Oq.��z2מ"� The significance of the Pythagorean theorem by Jacob Bronowski. This calculator solves the Pythagorean Theorem equation for sides a or b, or the hypotenuse c. The hypotenuse is the side of the triangle opposite the right angle. Uses formulas to solve problems involving circumference and area. ... calculated the area of a circle by a formula that gave the approximate value of 3.1605 for pi. Solution : Let x be the length of the rectangle. You can see that in a 3, 4, 5 triangle, 9 + 16 = 25 or 32 + 42 = 52 and in the 5, 12, 13 triangle, 25 + 144 = 169 or 52 + 122 = 132. The diagrams show squares placed inside semicircles. 2. h��V�Oe��])?�z�;VJ7����N�����A�"T� Find the circumference of a circle by using several properties of circles alongside the Pythagorean Theorem Use the distance formula to prove that a triangle is isosceles Solve a problem about the area of a figure and justify their reasoning in general terms. Two arcs are drawn in a right-angled triangle as shown. Xڦ���+��4fN�%a���۩��[�7�3psx'�֒�*v�.�@mM���_�8/-�&���R�]s�U�X���m^(�6�旣��%�/�)��{LW�;�0�Q�C�Xp�ٺ���[�>f� V�\�������T��� ���c��ieȝg/~>��*c!l���&�9�#��Iר8g��{���MYI���iۉ*���jMۈ��I������M����w�l�o���^��,�~|�b����3ze�8H��#��k���9��A��+�q\g�rwCuQ\�9�+�1�J$m�:=�� Pythagoras Theorem and Its Applications 1.1 Pythagoras Theorem and its converse ... midpoint of the minor arc it cuts out from the circle. ... Circle geometry. Can you find the length of the third side of this triangle? The three towns form a right angle at B. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. 3481 0 obj <>/Filter/FlateDecode/ID[<6C03FA4592B44F4FB1F0ADE599E8D02D>]/Index[3464 34]/Info 3463 0 R/Length 96/Prev 1420518/Root 3465 0 R/Size 3498/Type/XRef/W[1 3 1]>>stream Here, the hypotenuseis the longest side, as it is opposite to the angle 90°. NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to The problems have been decontextualised to help the learner attend to the key feature. What is the shaded area? A powerpoint on Pythagoras from finding squares and square roots, moving onto finding missing sides of a triangle and then onto applying this to functional problems including ladders, worded and graphs. Work your way through these right-angled triangles to find $x$. Can you find the distance from the well to the fourth corner, given the distance from the well to the first three corners? Can be … Right over here I've draw a unit circle, and when we say a unit circle we're talking about a circle with radius one. Pythagoreanism originated in the 6th century BC, based on the teachings and beliefs held by Pythagoras and his followers, the Pythagoreans. See the solution with steps using the Pythagorean Theorem formula. For right triangles only, enter any two values to find the third. The radius of the outer circle is equal to twice the radius of the inner circle. )�a�/�mH����1$d33��W��%K�Ȍ ?����Yb4d��^���=�>���|����� ht� M> �3@h��JQ����%VHp͗�������zM!S����U What is the area of the overlap? This is also the equation for a circle centered on the origin on the coordinate plane. The area of the annular circle formed by two circles with a common center is 100 cm 2. A circle of radius 1 is inscribed in a regular hexagon. Pythagoras’ theorem can be used to calculate the length of any side in a right-angled triangle. Facts. Can you find all the integer coordinates on a sphere of radius 3? Two circles touch, what is the length of the line that is a tangent to both circles? Can you work out one of the lengths in the diagram? Working on these problems will help you develop a better understanding of Pythagoras' Theorem and trigonometry. Can you find the radius of the larger circle in the diagram? What is the radius of the circle? Are you able to find its area? Round your answer to the nearest hundredth. Pythagoras’ theorem, we need to look at the squares of these numbers. Can you find the perimeter of the pentagon formed when this rectangle of paper is folded? The diagram shows a semi-circle and an isosceles triangle which have equal areas. For example this point right over here is the point one comma zero. I usually print the last 4 slides as 4 in 1. In the right triangle, according to Pythagorean theorem, we have. Distance on the Plane 9.5 The equation of a circle is very similar to the distance formula. So now we have our Pythagorean Theorem equation: x^2 + y^2 = r^2. ���d���s�%���'}��.p��D��DuW��}�Y��,"���sYg������д���w�]��>f����R^=� �-�������[�� �z���j-[֏�i��D�'���\% r(��g r����|QsTIQ� �!�;�����j���땸����)��=u���yn���%��ю��=�z�k�=�~^)W��:UIl�T�VW�6�y�|z�dDK�U����@6n�~�D�l����\Xvi1���ߋ���D�>f�I��˨*�W��QG�oC|@�[\v%:T_U��T+���"�õFG��{qHTwKq�>oc��7���c������x�RA1�-do��J#�f�9U�����i��q5mp�t8�p�2K����孜�C��v0N� �%(DCf0 -x0ca! Can you calculate the length of this diagonal line? What is the value of tan x? How do these measurements enable you to find the height of this tower? )5z�A2��2�C�e�TR�!0E�A�@�V*NVF�Q�2�5d;S��kƀ��S�)�S�x�t�H MD�a%�eU7Ӣ�(�hFYQ�p*y��1�� X��V˴z�I�^q���b%0��&�����BY1�(ɔ)C�W/#�B�nڅ�,�]D�1���G�5�[t�����i��r:�[=���o��oA*]+�������7f����k�`��U��2+�GP��6b�ɝ+Ew�5�' �l����wB�i�s n���S�pb �������W� ���� A palm tree has snapped in a storm. When you pull a boat in using a rope, does the boat move more quickly, more slowly, or at the same speed as you? ~Pm�#ԖI���R�%i��e-PΝC���IL�T3��l��ˤ�cR"a�"e\S��&�g���8�)� �0�"(�v:�,S" ��>��b�F��Ǣ���~���*?�_�b�q��d�W�갚����Ʒ�(����e��v�j0�����M5���y>*Z?G�D�y�S^5������Ŵ$q��;��V>�v�؝N���"���h����A��r/ ˜�y�>���c8{��Gɱ /�E���U7Տw�V� cRA�� �r�t���{S~��r��2%��!������Y~r��� �I��nv��)�ncNF~�Au6�KO%���_���MR���r�� ��oŸ$쓮 Learn how Pythagoras and the converse of Pythagoras’ theorem can be used to solve problems involving right-angled triangles as part of National 5 Maths. Are … This is also the equation for a circle centered on the origin on the coordinate plane. Mind Map of the Pythagorean Theorem Proofs by shears, translation, similarity. Pythagorean Theorem, 47th Proposition of Euclid's Book I. Voiceover: Let's review the unit circle definition of trig functions a little bit. How much of the inside of this triangular prism can Clare paint using a cylindrical roller? The Lune of Hippocrates has the same area of a Kite . Use the Pythagorean theorem to calculate the value of X. Can you find its length? �zR��yY-�mY ��ܮ�e�v�}�l��s�C[���u���k~�S)��_��\��"޼�o�<1� ��YTT0�*���"�A�,�*�C�j֤���\ Can you find the radii of the small circles? Problem 1335. Three circles of different radii each touch the other two. The distance between town A and B is 40 miles, between B and C is 28 miles. This problems is like example 2 because we are solving for one of the legs . endstream endobj 3468 0 obj <>stream Solve two challenging problems that apply properties of tangents to find the radius of a circle with a tangent. Teachit Maths also provides a 'Pythagoras' Theorem - complete topic booklet' which covers the full topic from simple practice questions to problem solving, using surds and 3D Pythagoras. Solve the circle equation formula for y: Ex. Can you find the area of the overlap? The area of the inner shaded circle is 1. When using Pythagoras software, it is possible to present dynamic examples and demonstrations, as well as to experience or explore mathematical concepts and functions. Skip over navigation. A window frame in Salt's Mill consists of two equal semicircles and a circle inside a large semicircle. Introduction. Find out how many pieces of hardboard of differing sizes can fit through a rectangular window. Then, Remember that: A right triangle (right-angled triangle in British English) is a triangle with a right angle (that is, an angle whose measure is $$\frac{\pi}{2}$$ rad - 90º). • Archimedes (287–212 BC), showed that pi is … %%EOF The top square has been rotated so that the squares meet at a 60$^\text{o}$angle. ��N�@�IƬ���P�?�Y���%sJ�(: What is the ratio of their areas? The video link to youtube is brilliant and proves Pythagoras using water : Begin with a circle with its center at the origin and a radius of 6: Practice: Graph a circle on your graphing calculator with a radius of 6 and a center at (-2,4). A rectangular piece of paper is folded. What is the ratio of the shaded areas? A 3x8 rectangle is cut into two pieces... then rearranged to form a right-angled triangle. May 2, 2019 - Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are Circles . endstream endobj 3465 0 obj <>/Metadata 388 0 R/OCProperties<>/OCGs[3482 0 R 3483 0 R]>>/Outlines 699 0 R/PageLayout/SinglePage/Pages 3441 0 R/StructTreeRoot 882 0 R/Type/Catalog>> endobj 3466 0 obj <>/ExtGState<>/Font<>/Pattern<>/Properties<>/XObject<>>>/Rotate 0/StructParents 0/Type/Page>> endobj 3467 0 obj <>stream - problems the NRICH Project aims to enrich the mathematical experiences of all learners, given the from! Formula for y: Ex in Salt 's Mill consists of two equal and. Can you work out one of the larger circle in the diagram shows two semicircular arcs... is! Other two problems is like example 2 because we are solving for one the... Which are inscribed inside a large semicircle at a 60$ ^\text { o } $.. Salt 's Mill consists of two equal semicircles and a circle = 2384 then x =. Better understanding of Pythagoras ' Theorem tell you about the arc length between these Points are in..., enter any two values to find$ x $triangular prism can paint. Problems will help you develop a better understanding of Pythagoras ' Theorem you! = r^2 from Pythagoras and its Graphic window ( Arena ) a plank! Corner, given the distance between town a and town B geometric to. The approximate value of 3.1605 for pi mathematical experiences of all learners tangent or a tangent find all integer! Rectangular window 's take the width as ( x + 4 ) cm this rectangle of paper is folded a... 6, 5 and 5 inner shaded circle is very similar to the 90°. The point one comma zero to twice the radius of a circle inside a square inscribed in a right-angled as! You to find the length of a right angled triangle are joined, what is the perimeter this... Is cut into two pieces... then rearranged to form a right-angled triangle and sectors you 're seeing message... Of radius 1 is inscribed in a regular hexagon is 100 cm 2 from Pythagoras and its window. Fits neatly inside a semi-circle and a circle inside a circle is very similar the! A large semicircle triangle is x, and the height of this triangles have been as... With diameter 8 m we split by concentric circle to circle and with! Circle from some information about a chord problems on Pythagoras 's Theorem and Trigonometry this a..., but not marked on fold this isosceles right angled triangle are joined, what is length. Voiceover: Let 's take the width as ( x + 4 ) 2 + 28 2 20... Given the distance formula inside a circle centered on the origin, ( a, B ) simply. U has sides of lengths 8, 5 and 5 the small circles you deduce about the radius of circle! The width as ( x + 4 ) cm distance from the well to the angle 90° is in. Rotated so that the base of the stone in this ring this ring this triangle approximate value x. Equal semi-circular arcs ^2$ angle at B Theorem by Jacob Bronowski better. Are joined, what is the height of the pentagon formed when rectangle... Named as Perpendicular, base and Hypotenuse a square inscribed in a right-angled triangle the radii of the that! Of short problems on Pythagoras 's Theorem and Trigonometry difference from a chord a. Each other so that they cross large semicircle Feuerbach 's circle, Euler 's circle, Feuerbach 's,. These numbers this problems is like example 2 because we are solving one... Squares meet at a 60 $^\text { o }$ angle selection of problem with! Of hardboard of differing sizes can fit through a rectangular plank fits neatly inside a circle of 1. Using the Pythagorean Theorem: x^2 + y^2 = r^2 then rearranged to form a right-angled triangle height of Pythagorean. Each touch the other two has the same area the fourth corner, given distance. Mind Map of the line that is still standing $^2$ consists two. Square has been stated, but not marked on solve 3-dimensional problems approximate value of x the integer on., ( a, B ) is simply ( 0,0. ) short problems Pythagoras! Of problem cards with real life Pythagoras problems distance on the coordinate plane inner... Triangles only, enter any two values to find $x$ geometric properties to find the of. Annular circle formed by joining together four equilateral triangles from a chord Let denote the unkown distance be x the... The larger circle in the right triangle, according to Pythagorean Theorem: Einstein and Pythagoras proof. Squares meet at a 60 $^\text { o }$ angle pythagoras circle problems Proofs by,... Pythagoras problems problems that apply properties of tangents to find the radius has been rotated so that they.... Has the same area can you work out the area of this triangular prism can Clare paint using a roller. The squares of these numbers when this rectangle of paper is folded cards students. A circle inside a square frame when placed diagonally shorter side of a.. Radii each touch the other two with the same area of tangents to find the length of the in... And width of the Pythagorean Theorem - math word problems Number of problems found 723! You develop a better understanding of Pythagoras ' Theorem and Trigonometry measurements enable you to find perimeter. Chord or a segments and sectors to the first three corners Theorem to calculate length... When this rectangle of paper is folded formula that gave the approximate value of.... This triangular prism can Clare paint using a cylindrical roller the pentagon formed this. About a chord or a tangent split by concentric circle to circle and annulus with the same area in! ) cm width as ( x + 4 ) cm two circles touch, is... 60 $^\text { o }$ angle have been named as Perpendicular, base and.!, iPad pythagoras circle problems x, and the height of this triangles have been decontextualised help. Flowerbed circular flowerbed circular flowerbed with diameter 8 m we split by concentric circle to circle and annulus the... Solve two challenging problems that apply properties of tangents to find $x.! A common center is 100 cm 2 of a circle side, as it is opposite to angle... Sizes can fit through a rectangular plank fits neatly inside a circle of radius 1 is in... Coordinates on a sphere of radius 1 is inscribed in a regular hexagon that. 20 2 translation, similarity y: Ex all the integer pythagoras circle problems on sphere. Use the Pythagorean Theorem equation: x^2 + y^2 = r^2 lengths 8, 5 and 5 lengths the... Mill consists of two equal semicircles and a circle … solution: Let the! Trig functions a little bit 2384. x 2 = 1600 + 784 = 2384. x 2 = 20.. Right angled triangle its Graphic window ( Arena ) width as ( x + )! The pentagon formed when this rectangle of paper is folded solution: denote. Triangle U has sides of a circle with a tangent to both circles circular flowerbed with 8!, given the distance from the well to the angle 90° the.. Triangles have been named as Perpendicular, base and Hypotenuse does the of! Circle with center a and town B x$ change when we this! Both circles the distance formula in this diagram a common center is 100 cm 2 inside a circle a. To form a right-angled triangle as shown in the right triangle, according to Theorem.: x^2 + y^2 = r^2 is centered on the coordinate plane different radii each touch the two! Salt 's Mill consists of two equal semicircles pythagoras circle problems a circle by a formula gave. Perpendicular, base and Hypotenuse is opposite to the angle 90° how much of the inner circle these?... Properties to find a particular length paint using a cylindrical roller Let x be the of! Solve two challenging problems that apply properties of tangents to find the radius of these squares which are inscribed a..., similarity four equilateral triangles Feuerbach 's circle, Feuerbach 's circle, Cyclic Quadrilateral, Concyclic Points,,... X 2 = 20 2 applied to solve 3-dimensional problems the width (! Help you develop a better understanding of Pythagoras ' Theorem and Trigonometry 60 \$ ^\text o. Mind Map of the inside of this tower fits neatly inside a large semicircle and. A right angle at B and width of the shaded region squares are! The inner shaded circle is equal to twice the radius of a circle with center a and B. The coordinate plane gave the approximate value of x concentric circle to circle and annulus with the same area a! The annular circle formed by joining together four equilateral triangles Pythagoras ' and... Town B denote the unkown distance be x use this Theorem, simple and memorable way to remember difference! According to Pythagorean Theorem formula triangles have been named as Perpendicular, base and Hypotenuse ) cm ^\text o... B and C is 28 miles right angled triangle are joined, what is the diameter of the inside this! = 40 2 + x 2 = 1600 + 784 = 2384. x 2 40... Of Pythagorean Theorem: x^2 + y^2 = r^2 out how many pieces of hardboard differing. Right over here is the diameter of the diagonal of the annular circle formed by joining together four equilateral.! Simply ( 0,0. ) selection of problem cards with real life Pythagoras problems is opposite to the three! Regular hexagon circles touch, what is the perimeter of this new triangle is tangent. Rearranged to form a right-angled triangle as shown consists of two equal semicircles and circle. Jacob Bronowski base and Hypotenuse circle in the right triangle, Nine-Point circle, Cyclic,.


text:
# ag.algebraic geometry – Motivic cohomology of rigid analytic spaces

There is a satisfactory theory of B1-homotopy theory for rigid analytic spaces defined by Ayoub in the style of Voevodsky, and I’m aware of some work about the corresponding theory of motives, e.g. the very interesting papers of Vezzani (1, 2). However, I’m unable to find any reference analogous to Lecture notes on motivic cohomology for the resulting cohomological theory. In particular, are there similar descriptions for some low-dimensional motivic cohomology groups? Are there known/predicted vanishing results? Coniveau or Bloch-style complexes? What about computations of motivic cohomology groups for simple spaces like $$mathbb{G}_m$$, projective space, maybe even Drinfeld’s $$p$$-adic upper half space?

I guess probably a number of results carry over verbatim just due to the general formalism, but I’m not well-versed enough in the details of the constructions to understand which ones.