# 5.7: Net Change - Mathematics

In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.

## Basic Integration Formulas

Recall the integration formulas given in [link] and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.

Example (PageIndex{1}): Integrating a Function Using the Power Rule

Use the power rule to integrate the function ( ∫^4_1sqrt{t}(1+t)dt).

Solution

The first step is to rewrite the function and simplify it so we can apply the power rule:

[ ∫^4_1sqrt{t}(1+t)dt=∫^4_1t^{1/2}(1+t)dt=∫^4_1(t^{1/2}+t^{3/2})dt.]

Now apply the power rule:

[ ∫^4_1(t^{1/2}+t^{3/2})dt=(frac{2}{3}t^{3/2}+frac{2}{5}t^{5/2})∣^4_1]

[ =[frac{2}{3}(4)^{3/2}+frac{2}{5}(4)^{5/2}]−[frac{2}{3}(1)^{3/2}+frac{2}{5}(1)^{5/2}]=frac{256}{15}.]

Exercise (PageIndex{1})

Find the definite integral of ( f(x)=x^2−3x) over the interval ([1,3].)

Hint

Follow the process from Example to solve the problem.

[ −frac{10}{3}]

## The Net Change Theorem

The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

Net Change Theorem

The new value of a changing quantity equals the initial value plus the integral of the rate of change:

[F(b)=F(a)+∫^b_aF'(x)dx]

or

[∫^b_aF'(x)dx=F(b)−F(a).]

Subtracting (F(a)) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.

We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure. Figure (PageIndex{1}): The graph shows speed versus time for the given motion of a car.

Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by

[ ∫^5_2v(t)dt=∫^4_240dt+∫^5_4−30dt=80−30=50.]

Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by

[ ∫^5_2|v(t)|dt=∫^4_240dt+∫^5_430dt=80+30=110.]

Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.

To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.

Example (PageIndex{2}): Finding Net Displacement

Given a velocity function (v(t)=3t−5) (in meters per second) for a particle in motion from time (t=0) to time (t=3,) find the net displacement of the particle.

Solution

Applying the net change theorem, we have

[ ∫^3_0(3t−5)dt=frac{3t^2}{2}−5t∣^3_0=[frac{3(3)^2}{2}−5(3)]−0=frac{27}{2}−15=frac{27}{2}−frac{30}{2}=−frac{3}{2}.]

The net displacement is ( −frac{3}{2}) m (Figure). Figure (PageIndex{2}): The graph shows velocity versus time for a particle moving with a linear velocity function.

Example (PageIndex{3}): Finding the Total Distance Traveled

Use Example to find the total distance traveled by a particle according to the velocity function (v(t)=3t−5) m/sec over a time interval ([0,3].)

Solution

The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.

To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,

(3t−5=0)

(3t=5)

( t=frac{5}{3}.)

The two subintervals are ( [0,frac{5}{3}]) and ( [frac{5}{3},3]). To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval ([0,frac{5}{3}]), we have (|v(t)|=−v(t)) over that interval. Over ([ frac{5}{3},3]), the function is positive, so (|v(t)|=v(t)). Thus, we have

( ∫^3_0|v(t)|dt=∫^{5/3}_0−v(t)dt+∫^3_{5/3}v(t)dt)

( =∫^{5/3}_05−3tdt+∫^3_{5/3}3t−5dt)

( =(5t−frac{3t^2}{2})∣^{5/3}_0+(frac{3t^2}{2}−5t)∣^3_{5/3})

( =[5(frac{5}{3})−frac{3(5/3)^2}{2}]−0+[frac{27}{2}−15]−[frac{3(5/3)^2}{2}−frac{25}{3}])

( =frac{25}{3}−frac{25}{6}+frac{27}{2}−15−frac{25}{6}+frac{25}{3}=frac{41}{6}).

So, the total distance traveled is ( frac{14}{6}) m.

Exercise (PageIndex{2})

Find the net displacement and total distance traveled in meters given the velocity function (f(t)=frac{1}{2}e^t−2) over the interval ([0,2]).

Hint

Follow the procedures from Example and Example. Note that (f(t)≤0) for (t≤ln4) and (f(t)≥0) for (t≥ln4).

Net displacement: ( frac{e^2−9}{2}≈−0.8055m;) total distance traveled: ( 4ln4−7.5+frac{e^2}{2}≈1.740 m)

## Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in Example.

Example (PageIndex{4}): How Many Gallons of Gasoline Are Consumed?

If the motor on a motorboat is started at (t=0) and the boat consumes gasoline at the rate of (5−t^3) gal/hr, how much gasoline is used in the first 2 hours?

Solution

Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [0,2]. We have

[ ∫^2_0(5−t^3)dt=(5t−frac{t^4}{4})∣^2_0=[5(2)−frac{(2)^4}{4}]−0=10−frac{16}{4}=6.]

Thus, the motorboat uses 6 gal of gas in 2 hours.

Example (PageIndex{5}): Chapter Opener: Iceboats

As we saw at the beginning of the chapter, top iceboat racers can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Figure (PageIndex{3}): (credit: modification of work by Carter Brown, Flickr)

Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function (v(t)=20t+5.) For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by

[ v(t)=egin{cases}20t+5& for 0≤t≤frac{1}{2}15 & for frac{1}{2}≤t≤1end{cases}.]

Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?

Solution

To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then

Distance =( ∫^1_02v(t)dt.)

Substituting the expressions we were given for (v(t)), we get

( ∫^1_02v(t)dt=∫^{1/2}_02v(t)dt+∫^1_{1/2}2v(t)dt)

( =∫^{1/2}_02(20t+5)dt+∫^1_{1/3}2(15)dt)

( =∫^{1/2}_0(40t+10)dt+∫^1_{1/2}30dt)

( =[20t^2+10t]|^{1/2}_0+[30t]|^1_{1/2})

( =(frac{20}{4}+5)−0+(30−15))

(=25.)

Andrew is 25 mi from his starting point after 1 hour.

Exercise (PageIndex{3})

Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function (v(t)=−10t+15.) In other words, the wind speed is given by

( v(t)=egin{cases}20t+5 & for 0≤t≤frac{1}{2}−10t+15& forfrac{1}{2}≤t≤1end{cases}).

Under these conditions, how far from his starting point is Andrew after 1 hour?

Hint

Don’t forget that Andrew’s iceboat moves twice as fast as the wind.

(17.5 mi)

## Integrating Even and Odd Functions

We saw in Functions and Graphs that an even function is a function in which (f(−x)=f(x)) for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are symmetric about the y-axis. An odd function is one in which (f(−x)=−f(x)) for all x in the domain, and the graph of the function is symmetric about the origin.

Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly ([−a,a],) evaluate to zero because the areas above and below the x-axis are equal.

Rule: Integrals of Even and Odd Functions

For continuous even functions such that (f(−x)=f(x),)

[∫^a_{−a}f(x)dx=2∫^a_0f(x)dx.]

For continuous odd functions such that (f(−x)=−f(x),)

[∫^a_{−a}f(x)dx=0.]

Example (PageIndex{6}): Integrating an Even Function

Integrate the even function ( ∫^2_{−2}(3x^8−2)dx) and verify that the integration formula for even functions holds.

Solution

The symmetry appears in the graphs in Figure. Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have

( ∫^2_{−2}(3x^8−2)dx=(frac{x^9}{3}−2x)∣^2_{−2})

( =[frac{(2)^9}{3}−2(2)]−[frac{(−2)^9}{3}−2(−2)])

( =(frac{512}{3}−4)−(−frac{512}{3}+4))

( =frac{1000}{3}).

To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.

[ ∫^2_0(3x^8−2)dx=(frac{x^9}{3}−2x)∣^2_0=frac{512}{3}−4=frac{500}{3}]

Since ( 2⋅frac{500}{3}=frac{1000}{3},) we have verified the formula for even functions in this particular example. Figure (PageIndex{4}): Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative area between the curve and the x-axis. Both views show the symmetry about the y-axis.

Example (PageIndex{7}): Integrating an Odd Function

Evaluate the definite integral of the odd function (−5sinx) over the interval ([−π,π].)

Solution

The graph is shown in Figure. We can see the symmetry about the origin by the positive area above the x-axis over ([−π,0]), and the negative area below the x-axis over ([0,π].) we have

[ ∫^π_{−π}−5sinxdx=−5(−cosx)|^π_{−π}=5cosx|^π_{−π}=[5cosπ]−[5cos(−π)]=−5−(−5)=0.] Figure (PageIndex{5}):The graph shows areas between a curve and the x-axis for an odd function.

Exercise (PageIndex{4})

Integrate the function ( ∫^2_{−2}x^4dx.)

Hint

Integrate an even function.

(dfrac{64}{5})

## Key Concepts

• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.

## Key Equations

• Net Change Theorem

( F(b)=F(a)+∫^b_aF'(x)dx) or (∫^b_aF'(x)dx=F(b)−F(a))

## Glossary

net change theorem
if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity

## 5.7: Net Change - Mathematics

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For additional MySQL 5.7 documentation, see the MySQL 5.7 Reference Manual, which includes an overview of features added in MySQL 5.7 (What Is New in MySQL 5.7), and discussion of upgrade issues that you may encounter for upgrades from MySQL 5.6 to MySQL 5.7 (Changes in MySQL 5.7).

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## Average Formulas

Average refers to the sum of numbers divided by n. Also called the mean average.

Sums of data divided by the number of items in the data will give the mean average. The mean average is used quite regularly to determine final math marks over a term or semester. Averages are often used in sports: batting averages which means number of hits to number of times at bat. Gas mileage is determined by using averages.

Hence, average = For Example: To find the average of 3, 5 and 7.

Step 1: Find the sum of the numbers.

Step 2: Calculate the total number.

Step 3: Finding average 15/3 = 5

### Quicker Method to Solve the Questions

Sum of elements = average × no. of elements

Example 1: The average of marks obtained by 4 students in a class is 65. Find the sum of marks obtained?
Solution. Here, number of marks obtained = 4

∴ sum of marks obtained = 65 × 4 = 260

Number of elements = Example 2: If the sum of elements and average are respectively 65 and 13, then find the number of elements.

Solution. Number of elements = = Average of a group consisting of two different groups whose averages are known:
Let a group with average a contain m quantities and another group of n quantities whose average is b, then the average of group c containing at a + b quantities Example 3: There are 30 students in a class. The average age of the first 10 students is 12.5 years. The average age of the next 20 students is 13.1 years. Find the average age of the whole class.

Solution.Total age of 10 students = 12.5 × 10 = 125 years

Total age of 20 students = 13.1 × 20 = 262 years

∴ Average age of 30 students = = = 12.9 years

If in a group one or more new quantities are added or excluded, then new quantity or sum.
= [change in no. of quantities × original average] ±
[change in average × final no. of quantities]
Take + ve sign if quantities added and take –ve sign if quantities removed

Example 4: The average weight of 24 students in a class is 35 kg. if the weight of the teacher is included, the average weight rises by 400 gms. find the weight of the teacher.

Solution.Total weight of 24 students = (24 × 35) kg = 840 kg

## Method 1: A Problem With a Decrease

Say one person weighed 150 pounds last year and now weighs 125 pounds. That's a decrease. The problem is to find the percent of decrease in weight (the weight loss).

First, subtract to find the amount of change:

150 - 125 = 25. The decrease is 25.

Next, divide the amount of change by the original amount:

Now, to change the decimal to a percent, multiply the number by 100:

The answer is 16.7%. So that's the percent of change, a decrease of 16.7% in body weight.

## Percentage change formula – mathematical examples

Let&aposs do a few examples together to get a good grasp on how to find percent change. In the first case, let&aposs suppose that you have a change in value from 60 to 72 and you want to know the percent change.

Firstly, you need to input 60 as the original value and 72 as the new value into the formula.

Secondly, you have to subtract 60 from 72 . As a result, you get 12 .

Next, you should get the absolute value of 60 . As 60 is a positive number, you don&apost need to do anything. You can erase the straight lines surrounding 60 .

Now, you can divide 12 by 60 . After this division, you get 0.2 .

The last thing to do is to multiply the 0.2 by 100 . As a result, you get 20 % . The whole calculations look like this:

[(72 – 60) / |60|] * 100 = (12 / |60|) * 100 = (12 / 60) * 100 = 0.2 * 100 = 20 %

You can check your result using the percentage change calculator. Is everything alright?

In the second example, let&aposs deal with a slightly different example and calculate the percent change in value from 50 to -22 .

Set 50 as the original value and -22 as the new value.

Then, you need to perform a subtraction. The difference between -22 and 50 is -72 . Remember always to subtract the original value from the new value!

Next, you are obliged to get the absolute of 50 . As the original value in this example is also a positive number, then you can just erase the straight lines.

It is time to perform the division. -72 divided by the 50 equals -1.44 .

Finally, you have to multiply the result by 100 . Let&aposs see. -1.44 times 100 is -144 % . The whole process should look like this:

[(-22 – 50) / |50|] *100 = (-72 / |50|) * 100 = (-72 / 50) * 100 = -1.44 * 100 = -144 %

Remember that you can always check the result with the percent change calculator.

In the third and final example, we will work only with negative numbers. In this case, you will see that getting the absolute value, may change the final result of an equation. We will find a percent change between -10 and -25 .

First, let&aposs assume that -10 is the original value that is being changed into -25 .

In the second step, as always, subtract the original value from the new one. -25 reduced by -10 is -15 .

Let&aposs concentrate during the third step, as it is different from what you have seen before. This time getting the absolute value will change something. Apply it to the original value -10 . As it is negative, you have to erase the minus before it, thus, creating a positive value of 10 . You will see that this change will have a significant effect on the final result.

Now, let&aposs divide -15 by 10 that you got from the last step. -15 divided by 10 is -1.5 .

You can finish your calculation by multiplying -1.5 by 100 . The final outcome is -150% . The full equation should look like this:

[(-25 – (-10)) / |-10|] * 100 = (-15 / |-10|) * 100 = (-15 / 10) * 100 = -1.5 * 100 = -150 %

As always, we encourage you to check this result with our percentage change calculator.

As you may have already observed, when the new value is smaller than the original one, the final result will be negative. Thus, you need to put a minus before it. On the other hand, if the new value is bigger than the original value, the result will be positive. You can use this to predict what the final result will be, and check your answer.

If you had used a negative instead of a positive for the absolute value in this example, then -15 would have been divided by -10 , giving you 1.5 as a result. It is a positive number, and your final answer would have been 150 % . Your error would have been the difference between -1.5 and 1.5 . This difference equals 3 , so our calculation would have ended with 300 % of an error ( 3 * 100% = 300 % )! This is why you have to be careful when solving mathematical problems. A small mistake in one place may result in an enormous error in another.

We have a task for you! Calculate, using the methods we have described previously, what is the percentage change between -20 and -30 . Concentrate and watch out for mathematical traps that are waiting for you. But don&apost fear. By this point, you should know everything that is required to do it correctly. Remember to check your result using the percent change calculator.

## MATH TESTS - Online Math Tests

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Know number names and the count sequence. Count to tell the number of objects. Identify and describe shapes.

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Use ratio reasoning to solve problems. Add, subtract, multiply and divide fractions. Compute fluently with multi-digit numbers and find common factors and multiples. Reason about and solve one-variable equations and inequalities.

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Use the order of operations to evaluate numeric expressions. Graph integers on a number line. Add, subtract, multiply, and divide rational numbers. Use divisibility rules to determine if a number is a factor of another number. Solve and graph equations and inequalities.

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## 5.7: Net Change - Mathematics

Prerequisite – Introduction of Set theory, Set Operations (Set theory)
For a given set S, Power set P(S) or 2^S represents the set containing all possible subsets of S as its elements. For example,
S = <1, 2, 3>
P(S) = <ɸ, <1="">, <2>, <3><1,2>, <1,3>, <2,3>, <1,2,3>>

Number of Elements in Power Set –
For a given set S with n elements, number of elements in P(S) is 2^n. As each element has two possibilities (present or absent>, possible subsets are 2×2×2.. n times = 2^n. Therefore, power set contains 2^n elements.

• Power set of a finite set is finite.
• Set S is an element of power set of S which can be written as S ɛ P(S).
• Empty Set ɸ is an element of power set of S which can be written as ɸ ɛ P(S).
• Empty set ɸ is subset of power set of S which can be written as ɸ ⊂ P(S).

Let us discuss the questions based on power set.

Q1. The cardinality of the power set of <0, 1, 2 . . ., 10>is _________.
(A) 1024
(B) 1023
(C) 2048
(D) 2043

Solution: The cardinality of a set is the number of elements contained. For a set S with n elements, its power set contains 2^n elements. For n = 11, size of power set is 2^11 = 2048.

Q2. For a set A, the power set of A is denoted by 2^A. If A = <5, <6>, <7>>, which of the following options are True.

(A) I and III only
(B) II and III only
(C) I, II and III only
(D) I, II and IV only

Explanation: The set A has 5, <6>, <7>has its elements. Therefore, the power set of A is:

Statement I is true as we can see ɸ is an element of 2^S.
Statement II is true as empty set ɸ is subset of every set.
Statement III is true as <5,<6>> is an element of 2^S.
However, statement IV is not true as <5,<6>> is an element of 2^S and not a subset.
Therefore, correct option is (C).

Q3. Let P(S) denotes the power set of set S. Which of the following is always true?

Solution: Let us assume set S =<1, 2>. Therefore P(S) = < ɸ, <1>, <2>, <1,2>>
Option (a) is false as P(S) has 2^2 = 4 elements and P(P(S)) has 2^4 = 16 elements and they are not equivalent.
Option (b) is true as intersection of S and P(S) is empty set.
Option (c) is false as intersection of S and P(S) is empty set.
Option (d) is false as S is an element of P(S).

Countable set and its power set –
A set is called countable when its element can be counted. A countable set can be finite or infinite.
For example, set S1 = representing vowels is a countably finite set. However, S2 = <1, 2, 3……>representing set of natural numbers is a countably infinite set.

• Power set of countably finite set is finite and hence countable.
For example, set S1 representing vowels has 5 elements and its power set contains 2^5 = 32 elements. Therefore, it is finite and hence countable.
• Power set of countably infinite set is uncountable.
For example, set S2 representing set of natural numbers is countably infinite. However, its power set is uncountable.

Uncountable set and its power set –
A set is called uncountable when its element can’t be counted. An uncountable set can be always infinite.
For example, set S3 containing all fractional numbers between 1 and 10 is uncountable.

• Power set of uncountable set is always uncountable.
For example, set S3 representing all fractional numbers between 1 and 10 is uncountable. Therefore, power set of uncountable set is also uncountable.

Let us discuss gate questions on this.

Q4. Let ∑ be a finite non-empty alphabet and let 2^∑* be the power set of ∑*. Which of the following is true?

Solution: Let ∑ =
then ∑* = < ε, a, b, aa, ba, bb, ……………….>.
As we can see, ∑* is countably infinite and hence countable. But power set of countably infinite set is uncountable.
Therefore, 2^∑* is uncountable. So, the correct option is (C).

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## Calculate the Distance Between Two Points

We use what's known as the distance formula to obtain the distance between any two points in space. Suppose you need to determine the distance between the points (2,0) and (5,0). Here's how they look on a graph: With just a quick glance, you can tell they are 3 units apart. They're both on the x axis, so it's just a straight line measurement. In fact, you may not have even needed the graph at all to figure that one out. But, how can you measure the distance between two random points on a graph? Why, use the distance formula of course!

### Breaking it down

At first glance, that formula looks like a real mess! But just think about the (x_<2>-x_<1>) and (y_<2>-y_<1>) components as the length in each direction. Using those two values we mathematically construct an imaginary triangle with two legs whose length we can calculate. Let me show you what I mean visually: We're just measuring the distance along each axis, and then using the pythagorean theorem to compute the length of the hypotenuse, which is the imaginary line directly between our two points. It doesn't matter which point is ((x_<1>,y_<1>)) and which is ((x_<2>,y_<2>)). The key idea to take away from this plot is that you only care about the change in x and the change in y. You'll use each of those measurements as the side of a triangle, where the hypotenuse is the distance between the two points. You could express the distance formula like this:

How did we come up with that formula? It's simply the pythagorean theorem, which allows us to find the hypotenuse of a right triangle. In our case, the hypotenuse is the distance between the two points!

Let's look at a simple example now:

### Example:

Find the distance between the two points (5,5) and (1,2) by using the distance formula.

Rather than blindly plugging numbers into a formula, draw a graph so you know what's going on. Those are your two points. Remember that the important values are the change in x and the change in y. That's equivalent to saying, what is (x_2-x_1) and what is (y_2-y_1).

Here, the change in x is 4 units, and the change in y is 3 units. You can even count the lines on the graph to make sure that's correct.

Once we knew that (x_<2>-x_<1>) was 4 and (y_<2>-y_<1>) was 3, we just inserted those numbers into the distance formula to solve.

You might be wondering what happens if you reverse the points? Remember that we are dealing with distances, which are inherently positive. The distance is the same in either direction, going from point 1 to point 2 or vice versa. So, just use the positive distance between two points. If you look at the formula, you'll notice that (x_<2>-x_<1>) and (y_<2>-y_<1>) are squared, which automatically makes them positive anyway. So it doesn't matter which point is which -- you'll still get a positive distance!

Let's look at another example and solve it without using a graph:

### Example:

Find the distance between the two points (8,-2) and (3, 9).

Now that you understand how the distance formula works, you can plug the numbers straight into the formula:

The distance formula is not complicated -- you just need to practice with a graph so you understand what's going on. You're simply making a triangle and finding the length of the hypotenuse. The distance formula is just the pythagorean theorem!

### Extending to three-dimensions

What about a three-dimensional space? How would we find the distance between the points (1,5,0) and (2,0, 8)? It's certainly a lot harder to draw a graph and measure the distance! The three-dimensional distance formula is actually super simple. Just add another term into the formula because we have to account for the z-axis now.

If you've made it this far, hopefully you have a better understanding of the distance formula. If that's not the case, here's another distance formula lesson available online.

#### Summary

The distance formula computes the distance between two coordinate points.

## Elementary Math Games

Kids love to play games, and they can often learn so much through playing. If you are looking for free elementary math games that your children or students can play online, then you have come to the right place.

The games are organized by topic and grade level.

Here you can find fun games suitable for first grade students.

Second grade kids can learn so much by playing these interactive math games.

In 3rd grade, kids learn many new concepts. Drills and repetitions are important, but boring. These games make drills exciting and fun.

Let your students practice important math concepts by playing these cool games.

Rounding Numbers Pirate Game
Round whole numbers and decimals in this fun pirate math game.

Adding whole numbers, decimals, fractions, and signed numbers.

Subtraction Games
Subtracting whole numbers, decimals, fractions, and integers.

Multiplication Math Games
Multiplying one-digit, two-digit, and larger whole numbers, as well as decimals, fractions, and integers.

Place Value Games
Kids will learn the place value in whole numbers and decimals.

Counting to 5 Pirate Game
Play this fun game to practice counting to 5.

Money Games
Learn about counting money and change, sales tax, simple and compound interest.

Decimal Games
Add, subtract, multiply, and divide decimals. Change them to fractions or percents.

Fraction Games
Add, subtract, multiply, and divide decimals. Change them to decimals or percents.

Geometry Games
Classify different plane and three dimensional geometric figures.

Math Vocabulary Games
These games are designed to reinforce important math vocabulary terms and definitions for K-8 math content.

## 5.7: Net Change - Mathematics

Dr. Raj Shah explains why math is taught differently than it was in the past and helps address parents' misconceptions about the "new math."
Watch the video

### Everyday Mathematics Research and Results

Everyday Mathematics is grounded in an extensive body of research into how students learn.

### Everyday Mathematics Implementation Measurement

Implementation measurement provides information about how teachers are implementing EM materials, what is actually happening in schools, classrooms, and districts “on the ground,” and why.

### Everyday Mathematics and the Common Core State Standards for Mathematical Practice

Andy Isaacs, director of EM revisions, discusses the CCSSM edition of Everyday Mathematics . Learn more

Professional Development Everyday Mathematics 4 Summer Offerings

#### Everyday Mathematics 4 Professional Development Leader Academy (PDLA) Level 1 (formerly Train the Trainer)

Attention EM4 teachers! We are offering FREE unit planning webinars for Grades 3 to 6 throughout the school year. EM4 authors at the University of Chicago will walk you through each unit and allow you to plan and ask questions.

Unit planning webinars for Kindergarten and Grades 1 and 2 were held last year. You can find recordings of those webinars on the VLC Resources page.

The Everyday Mathematics curriculum is produced by the UChicago STEM Education and the University of Chicago School Mathematics Project.

### Professional Development Everyday Mathematics 4 New User Training

CEMSE is pleased to announce that it will be offering professional development sessions this July for new users of Everyday Mathematics 4 . Learn more or register here.

### Professional Development Summer Everyday Math User's Conference

The Everyday Mathematics online professional development modules are now open for registration. Please visit the Online PD page on the Everyday Mathematics Virtual Learning Community for more information. Learn more

Find out more about the EM New-User Workshops offered by CEMSE over the 2012-2013 school year.

Browse through questions and answers, sorted by topic, Everyday Mathematics teachers have had about the curriculum throughout the years.

Find out more about the EM New-User Workshops offered by CEMSE on Saturday, April 21, 2012.

Find out how the EM curriculum meets the CCSS-M standards and how to integrate the standards into EM instruction. Learn more

Andy Isaacs, director of EM revisions, discusses the CCSSM edition of Everyday Mathematics . Learn more

Correlations between EM goals and the Standards for Mathematical Practice. How to integrate standards into EM instruction. Join the Virtual Learning Community to access EM lesson videos from real classrooms, share EM resources, discuss EM topics with other educators, and more.

The Crosswalk provides information about how the 2007 edition of Everyday Mathematics has been updated to meet the content requirements of the Common Core State Standards for Mathematics.

## Help with Fractions

### Fractions

/ To enter a fraction of the form 3/4. Click a number and then click fraction bar, then click another number.

↔ You can use fraction space button to create a number of the form 5 3/4. Enter a number, then click fraction space, click another number and then click on the fraction bar button, lastly enter another number.

DEC FRA Decimal format button and Fraction format button work as pair. When you choose the one the other is switched off.
Decimal format button is used for all decimal work. Also to change a fraction of the form 3/4 to the decimal 0.75, or a fraction of the form 7/4 or a mixed number of the form 1 3/4 to the decimal 1.75. Click on the decimal format button, enter a fraction or mixed number, then click equals. If the fraction or mixed number is only part of the calculation then omit clicking equals and continue with the calculation per usual. i.e. 3/4 DEC x 6 =.
Fraction format button is used to work with all fractions. Also to change a decimal of the form 0.5 to the fraction 1/2, or change a decimal of the form 1.75 to a mixed number of the form 1 3/4 or to the fraction 7/4, or a fraction of the form 7/4 to the mixed number 1 3/4. Click the fraction format button, enter a decimal, click equals and then click on a fraction form and then click equals. If the fraction of decimal is part of a calculation, omit clicking equals and continue with the calculation.

a b / c a+b / c Proper fraction button and Improper fraction button work as pair. When you choose the one the other is switched off.
Proper fraction button is used to change a number of the form of 9/5 to the form of 1 4/5. A proper fraction is a fraction where the numerator (top number) is less than the denominator (bottom number).
Improper fraction button is used to change a number of the form of 1 4/5 to the form of 9/5. An improper fraction is a fraction where the numerator (top number is greater than or equal to the denominator (bottom number).