We have the expression:
[tex]3(\frac{7}{5}x+4)-2(\frac{3}{2}-\frac{5}{4}x)[/tex]We simplify it as follows:
[tex]\frac{21}{5}x+12-3+\frac{5}{2}x\Rightarrow(\frac{21}{5}x+\frac{5}{2}x)+(12-3)[/tex][tex]\Rightarrow\frac{67}{10}x+9[/tex]From this, we have that the solution is the B option.
Question 8 of 10Jerry drew AJKL and AMP so that < K =¿N, LL = LP, JK= 6, andMN = 18. Are A JKL and A MNP similar? If so, identify the similarity postulateor theorem that applies.
Solution.
Given
In triangle JKL and MNP,
Thus, triangle JKL and MNP are equiangular
Hence, we can conclude that both triangles are similar by AA
The answer is option A
Answer:
A
Step-by-step explanation:
The function, fx) = x^2 - 4x + 3, has y-values that increase when x<2. TrueFalse
Let's begin by listing out the information given to us:
[tex]\begin{gathered} f(x)=x^2-4x+3 \\ f(x)=y \\ \Rightarrow y=x^2-4x+3 \\ y=x^2-4x+3 \end{gathered}[/tex]We will proceed to choose values for x (values of x lesser than 2); x = 1, 0, -1
[tex]\begin{gathered} y=x^2-4x+3 \\ x=1 \\ y=1^2-4(1)+3=1-4+3=4-4=0 \\ y=0 \\ (x,y)=(1,0) \\ \\ x=0 \\ y=0^2-4(0)+3=0-0+3=3 \\ y=3 \\ (x,y)=(0,3) \\ \\ x=-1 \\ y=(-1^2)-4(-1)+3=1+4+3=8 \\ y=8 \\ (x,y)=(-1,8) \end{gathered}[/tex]From the calculation, we see a trend that the y-values increase as the x-value decreases. Hence, it is true
Instructions: Given the recursive rule, match it to the explicit form.
Explanation:
If we have a recursive expression with the form
[tex]a_n=a_{n-1}\cdot c[/tex]Then, the explicit formula is
[tex]a_n=a_1\cdot c^{n-1}[/tex]Therefore, for each option, we get:
[tex]\begin{gathered} a_n=a_{n-1}\cdot2\text{ with a}_1=1 \\ \text{ Then} \\ a_n=1\cdot2^{n-1}=2^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot-2\text{ with a}_1=2 \\ \text{ Then} \\ a_n=2\cdot(-3)^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot4\text{ with a}_1=-1 \\ \text{ Then} \\ a_n=-1\cdot4^{n-1}=-4^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot2\text{ with a}_1=-3 \\ \text{ Then} \\ a_n=-3\cdot2^{n-1} \end{gathered}[/tex]Answer:
Therefore, the answer is:
Find the slope of the linear function f with f(2) = 16 and f(4) = -2
f(2) = 16, Let this be represented as (2, 16)
f(4) = -2, Let this be represented as (4, -2)
[tex]\begin{gathered} \text{slope =}\frac{change\text{ in y}}{\text{change in x}} \\ \\ \text{slope = }\frac{-2-16}{4-2} \\ \text{slope = }\frac{-18}{2} \\ \\ \text{slope}=\text{ -9} \end{gathered}[/tex]Find the solution of the system of equations.5 +2g = 185x-Y=36
Ok we have the following system of equations:
[tex]\begin{gathered} 5x+2y=18 \\ 5x-y=36 \end{gathered}[/tex]So the first thing to do is take one of the equations above and clear either x or y. I'm going to pick the second equation and clear y:
[tex]\begin{gathered} 5x-y=36 \\ 5x=36+y \\ 5x-36=y \\ y=5x-36 \end{gathered}[/tex]Now we substitute this result in the first equation:
[tex]\begin{gathered} 5x+2y=5x+2\cdot(5x-36)=18 \\ 5x+10x-72=18 \\ 15x=18+72=90 \\ x=\frac{90}{15}=6 \end{gathered}[/tex]Now that we know x we take the result of clearing y from the second equation and find its value:
[tex]\begin{gathered} y=5x-36 \\ y=5\cdot6-36=30-36 \\ y=-6 \end{gathered}[/tex]So in the end x=6 and y=-6.
Task: Find the value of x and y that proves these triangles congruent. Instructions In one part you will find the value of x that proves the triangles congruent. In the second part you will find the value ofy that proves the triangles congruent. (G.6) (2 point) Complete each of the 2 activities for this Task. Activity 1 of 2 Find the value of x.(G.6)(1 point) 24 HI 31 7x-4 to 4y+8
Activity 1:
We are given two triangles. The two side lengths of one triangle are known but of the other are not. Our task is to choose the value of x and y that will make the triangles congruent.
Now, the side lengths that are congruent are with 31 in the rightmost triangle and 7x -4 in the left-most triangle; therefore, equating them gives
[tex]7x-4=31[/tex]Similarly, side length 24 must equal 4y+8; therefore,
[tex]4y+8=24[/tex]Now we have to choose the values of x and y that will make both equations true.
Let us solve for x in the first equation by first adding 4 to both sides. Doing this gives
[tex]7x=35[/tex]Finally, dividing both sides by 7 gives
[tex]x=5.[/tex]Activity 2:
Now, for the value of y.
To solve for y, we first subtract 8 from both sides to get
[tex]4y=16[/tex]Finally, dividing both sides by 4 gives
[tex]y=4.[/tex]Hence, to conclude x = 5 and y = 4.
In the figure below, c || d. Classify each of the following angle pairs, and tell whether they arecongruent or supplementary.6. 21 and 23Supp.7. 26 and 23Supp.8. 21 and 283/47/8Supp.5/6t9. 27 and 24Supp.10. 22 and 21& Supp.
Two angles are congruent if they are equal and supplementary if there sum is 180.
Given data:
c and d are parallel.
Now
[tex]\angle1,\angle3[/tex]are corresponding angles, so they are equal.
So,
[tex]\angle1,\angle3[/tex]form a congruent pair.
Now since
[tex]\angle1=\angle6[/tex]since they are vertically oppsoye angles.
And,
[tex]\angle1=\angle3(\text{corresponding angles)}[/tex]So,
[tex]\angle6=\angle3[/tex]So,
[tex]\angle6,\angle3[/tex]form a congruent pair.
Now,
[tex]\begin{gathered} \angle3=\angle8(vertically\text{ opposite angles)} \\ \angle1=\angle3(corresponding\text{ angles)} \\ \Rightarrow\angle1=\angle8 \end{gathered}[/tex]So,
[tex]\angle1,\angle8[/tex]form a congruent pair.
[tex]\angle7=\angle4(vertically\text{ opposite angles)}[/tex]So,
[tex]\angle7,\angle4[/tex]form a congruent pair.
Now,
[tex]\angle1+\angle2=180(linear\text{ pair)}[/tex]So,
[tex]\angle1,\angle2[/tex]form a supplementary pair.
The rate of change of function "f" is the same from x = -2 to x = 1 as it is from x = 1 to x = 4.Function “f” is a ? function
Linear functions have a constant rate of change for any interval of x-values. Then, function “f” is a linear function
Evaluate the expression b= 3/10c= 2/153c-bwrite in the simplest form
Evaluate those values into the expression:
[tex]\begin{gathered} 3c-b \\ so\colon \\ 3(-\frac{2}{15})-\frac{3}{10} \\ -\frac{6}{15}-\frac{3}{10}=\frac{-60-45}{150}=-\frac{105}{150}=-\frac{7}{10} \\ \end{gathered}[/tex]Answer:
[tex]-\frac{7}{10}[/tex]Answer:
-7/10
Step-by-step explanation:
3 × -2 / 15 - 3/10
-12/30 - 9/30
-21/30
-7/10
Which formula is used to determine the standard normal random variable (Z)?
The standard normal random variable Z can be calculated using the formula:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]Where x is the input, μ is the mean and σ is the standard deviation.
Therefore the correct option is the first one.
Part A: which of the following can be used to find the measure of angles
We shall begin by calculating the value of x, as that would help us to find the angle measure of each angle.
[tex]\begin{gathered} \angle C+\angle D+\angle E=180 \\ x-5+2x-3+x=180 \\ 4x-8=180 \\ 4x=180+8 \\ 4x=188 \\ x=\frac{188}{4} \\ x=47 \\ \angle C=x-5 \\ \angle C=47-5 \\ \angle C=42 \\ \angle D=2x-3 \\ \angle D=2(47)-3 \\ \angle D=94-3 \\ \angle D=91 \\ \angle E=x \\ \angle E=47 \end{gathered}[/tex]The triangle is a scalene triangle (all angles are different in measure)
(1) Part A; we can find the angles using the Triangle angle-sum theorem
(2) Part B; measure of each angle as shown as;
The angles are;
C = 42
D = 91
E = 47
Find the sum: (502 + 8d + )+(502 + 3d + 4)
The sum:
(502 + 8d ) + ( 502 + 3d + 4 )
Clearing the brackets, we get,
502 + 8d + 502 + 3d + 4
Collecting the like terms, we get,
8d + 3d + 502 + 502 + 4
11d + 1008
The correct answer: 11d + 1008
Write an equation in slope-intercept form for the line that is perpendicular to y = 3x + 7 and passes through the point (-6, 9).
y = -x/3 +11 is the line perpendicular to y = 3x +7 and passes through the point (-6,9)
What is a slope-intercept form?It gives the graph of a straight line and it is represented in the form
y= mx +c. It is one of the form used to calculate the equation of a straight line. We have to calculate the slope of the line from the equation. The slope calculated can be used in the slope-intercept form. It is the most popular form of a straight line.
We need to find the perpendicular slope to the line y = 3x +7.
The slope of a line perpendicular to m is -1/m
Here, from the equation y=3x+7, m=3
So,-1/m = -1/3
The slope-intercept form is,
y-y1=m(x-x1)
y - 9 = -1/3 * (x+6)
Now, simplify the above equation
y-9= -x/3 +6/3
By adding 9 on both sides, we get
y= -x/3 +11
y = -x/3 +11 is the line perpendicular to y = 3x +7 and passes through the point (-6,9)
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Divide 1/4 ÷ 2/3 and express the answer in simplest terms.
Given the expression
1/4 ÷ 2/3
This is expressed as 1/4 * 3/2
multiply the numerator and denominator together to have;
1/4 * 3/2
= (1*3)/(4*2)
= 3/8
Hence the expression in its simplest form is 3/8
a 2 ft by 2 ft square is divided into smaller squares and portions are shaded. What is the are of the portion and shades portion.?
Answer:
[tex]1.5ft^2[/tex]Explanation:
Here, we want to get the area of the shaded portion
To get this, we need the entire area
The entire area would be the product of the sides of the big square:
[tex]2\text{ }\times2=4ft^2[/tex]Now, let us count the number of shaded small squares.
6 out of 16 squares are shaded
The area of the shaded porion is thus:
[tex]\frac{6}{16}\times4=1.5ft^2[/tex]What is a multiple root of a polynomial and how do you find it?
The Solution:
The given polynomial is
[tex]P(x)=2x^4-4x^3-16x^2[/tex]A root of the polynomial P(x) is the value of x for which the polynomial P(x) is equal to zero.
That is, any value of x that makes P(x) = 0, is a root of P(x).
The Multiplicity of a Root: This is the number of times a particular root appears as a root in a polynomial.
To find the root of a polynomial, say P(x), you have to equate P(x) to zero, and then solve for the value of x.
So, we shall follow the above procedures to find the root(s) of P(x), and thereafter determine if there are multiple roots.
[tex]\begin{gathered} P(x)=2x^4-4x^3-16x^2=0 \\ \text{Factoring out 2x}^2,\text{ we have} \\ 2x^2(x^2-2x-8)=0 \end{gathered}[/tex]This means that:
[tex]\begin{gathered} x^2-2x-8=0 \\ or \\ 2x^2=0 \end{gathered}[/tex]Solving quadratic equations above by Tthe Factorization Method, we get
[tex]\begin{gathered} x^2-2x-8=0 \\ x^2-4x+2x-8=0 \\ x(x-4)+2(x-4)=0 \\ (x-4)(x+2)=0 \end{gathered}[/tex]So,
[tex]\begin{gathered} P(x)=2x^2(x-4)(x+2)=0 \\ \text{This means} \\ 2x^2=0\text{ }\Rightarrow x=0 \\ x-4=0\text{ }\Rightarrow x=4 \\ x+2=0\text{ }\Rightarrow x=-2 \\ So,\text{ the roots of P(x) are 0, -2, and 4} \end{gathered}[/tex]Looking at the roots of P(x) above, there is no root that appears more than once, hence, the multiplicity of each of the roots is one.
4) A cannonball is shot out of a cannon at a 459angle with an approximatecannon from which the ball was fired sits on the edge of a cliff, and its he20 meters. The equations given below represent the cannonball's heighand its horizontal distance (x) from the face of the cliff, (E)seconds afterHow many seconds after the ball was fired does its verticat height abovehorizontal distance from the cliff?
Let t be the amunt of seconds that have passed when the height of the cannonball above the ground is the same as its horizontal fistance from the cliff.
Since the height of the cannonball above the ground is represented using the variable y and the horizontal distance from the cliff is represented using the variable x, then, the condition that the height equals the horizontal distance can be expressed as:
[tex]y=x[/tex]Replace the expressions for y and x in terms of t into the equation:
[tex]-5t^2+2t+20=2t[/tex]We obtained a quadratic equation on the variable t.
Notice that the term 2t appears in both members of the equation. Then, it can be cancelled out:
[tex]-5t^2+20=0[/tex]Solve for t²:
[tex]\begin{gathered} \Rightarrow-5t^2=-20_{} \\ \Rightarrow t^2=\frac{-20}{-5} \\ \Rightarrow t^2=4 \end{gathered}[/tex]Take the square root to solve for t:
[tex]\begin{gathered} \Rightarrow t=\pm\sqrt[]{4} \\ =\pm2 \end{gathered}[/tex]Since t must be greater or equal to 0, then the negative solution should be discarded.
Therefore, the vertical height of the cannonball equals its horizontal distance from the cliff 2 seconds after the ball is fired.
The correct choice is option B) 2
If a triangle ABC is at: A = ( 2, 9 ) B = ( 5, 1 ) C = ( - 6, - 8 ) and if it is translated right 2 and down 7, find the new point B'.
Solution
Step 1
Triangle ABC is at: A = ( 2, 9 ) B = ( 5, 1 ) C = ( - 6, - 8 )
Step 2
If it is translated right 2 and down 7
B = (5, 1)
B' = ( 5+2, 1-7)
B' = ( 7, -6)
Final answer
B' = ( 7, -6)
determine whether the given below each equation represents a direct variation or not if it does find the constant of the variation 3y = 4x
,y= kx
Where k is the constant of variation.
First solve for y:
3y=4x
y= 4/3 x
So, the constant of variation k= 4/3
It represents a direct variation.
Craig like to collect vinyl records. Last year he ahead 10 records in his collection. Now he has 12 records. What is the percent increase?
Last year, Craig had 10 records.
Now, he has 12 records.
What is the percent increase?
The percent increase is given by
[tex]\%\: increase=\frac{\text{new value-old value}}{\text{old value}}\times100[/tex]In this case,
Old value = 10 records
New value = 12 records
[tex]\begin{gathered} \%\: increase=\frac{\text{new value-old value}}{\text{old value}}\times100 \\ \%\: increase=\frac{12-10}{10}\times100 \\ \%\: increase=\frac{2}{10}\times100 \\ \%\: increase=20 \end{gathered}[/tex]Therefore, there is a 20% increase in his record collection.
Yesterday, all three restaurants sold the number of meals that resulted in them earning the maximumprofit.Put the restaurants in order from least to most profit earned.Drag each tile to the correct box.
For Franco's Hotdogs, the graph shows a parabola. The profit is the y-value of the function, as can be seen, the maximum profit earned is $200 for Franco's Hotdogs.
Now, for Hanna's Barbeque the maximum profit earned is the maximum h(x) value shown in the table, then the maximum profit is $250 for Hanna's Barbeque.
For Rhonda's Burgers, it says the maximum profit is $227.
Then, the restaurant with the least profit earned is Franco's Hotdogs, the next one is Rhonda's Burgers and the restaurant with the most profit earned is Hanna's Barbeque, because:
[tex]200<227<250[/tex]Thus, that is the order.
Answer:
Franco's Hotdogs, Rhonda's Burgers, Hannah's Barbeque
Step-by-step explanation:
I did the tutorial
what are the roots of the quadratic equation below?[tex]3 {x}^{2} + 9x - 2 = 0[/tex]
Given:
A quadratic equation is:
[tex]3x^2+9x-2=0[/tex]Find-:
The roots of the quadratic equation
Explanation-:
Use quadratic formula:
[tex]ax^2+bx+c=0[/tex]Roots of the equation,
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]The roots of the given equation are:
[tex]3x^2+9x-2=0[/tex][tex]\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{9^2-4(3)(-2)}}{2(3)} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{81-(-24)}}{6} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{81+24}}{6} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{105}}{6} \end{gathered}[/tex]The roots of a quadratic equation are:
[tex]\begin{gathered} x_{1,2}=\frac{-9\pm\sqrt{105}}{6} \\ \\ x_1=\frac{-9+\sqrt{105}}{6},x_2=\frac{-9-\sqrt{105}}{6} \end{gathered}[/tex][tex]\begin{gathered} x_1=\frac{-9+\sqrt{105}}{6},x_2=\frac{-9-\sqrt{105}}{6} \\ \\ x_1=0.2078,x_2=-3.2078 \end{gathered}[/tex]The roots of a quadratic equation are 0.2078 and -3.2078.
QuestionsWhat is the equation of the line?y = 2x - 4y = 1/2x + 2y = 2x + 2y = 1/2x-4
We have the graph of the equation, and we want to know the equation of the line.
We remember that we need to parts: the slope and the y-intercept. On the graph, we see that when x=0, the graph passes through the point 2, and thus the y-intercept is 2.
[tex]b=2[/tex]Lastly, we will find the slope. For doing so, we will find two values of the line. In this case, we saw that the y-intercept is 2, so a point is (0,2).
Other point is (-4,0), as the x-intercept is -4.
For finding the slope, we remember the formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{0-(-4)}=\frac{2}{4}=\frac{1}{2}[/tex]And thus, the slope is 1/2.
This means that the line equation will be:
[tex]\begin{gathered} y=mx+b \\ y=\frac{1}{2}x+2 \end{gathered}[/tex]
Let R be the event that a randomly chosen athlete runs. Let W be the event that a randomly chosen athlete lifts weights.Identify the answer which expresses the following with correct notation: The probability that a randomly chosen athlete liftsweights, given that the athlete runs.
vertices abc are a(-4,5), b(-2,4), c(-3,2) if abc is reflected across the line y= -2 to produce the image abc; find the coordinates of vertex A
So, the coordinates of the new vertex A must be
[tex](-4,5-14)=(-4,-9)[/tex]So, the coordinates of vertex A is (-4,-9)
CAN SOMEONE HELP WITH THIS QUESTION?✨
Step-by-step explanation:
as this is a right-angled triangle, we use Pythagoras to get also c :
c² = a² + b² = 2² + 7² = 4 + 49 = 53
c = sqrt(53)
we know, sine = opposite/Hypotenuse.
so,
sin(A) = 2/sqrt(53) = 0.274721128...
from the norm circle we know cosine is the other leg of the right-angled triangle :
cos(A) = 7/sqrt(53) = 0.961523948...
tan(A) = sin(A)/cos(A) = 2/7 = 0.285714286...
sec(A) = 1/cos(A) = sqrt(53)/7 = 1.040015698...
csc(A) = 1/sin(A) = sqrt(53)/2 = 3.640054945...
cot(A) = 1/tan(A) = cos(A)/sin(A) = 7/2 = 3.50
oh, and FYI :
A = 15.9453959...°
3.8 times 24 long multipilcalion
please help I can't get no more wrong I 5,8 j 9,8 H 5,3
The coordinates of the points are:
H (5, 3)
I (5, 8)
J (9, 8)
Given that H and I have the same x-coordinate, then the side length of side HI is obtained, subtracting the y-coordinates, as follows:
[tex]HI=y_I-y_H=8-3=5[/tex]Given that I and J have the same y-coordinate, then the side length of side IJ is obtained, subtracting the x-coordinates, as follows:
[tex]IJ=x_J-x_I=9-5=4[/tex]The side length between H and I is 5 units
The side length between I and J is 4 units
Solve the quadratic equation by completing the square.x ^ 2 - 18x + 70 = 0 First, choose the appropriate form and fill in the blanks with the correct numbers. Then, solve the equation. Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas.
Answer:
Form:
[tex]\boxed{(x-9)^2=11}[/tex]Solution:
[tex]x=12.32,5.68[/tex]Explanation:
Step 1. The expression we have is:
[tex]x^2-18x+70=0[/tex]And we are required to find the appropriate form after completing the square, and then the solution or solutions to the equation.
Step 2. Compare the given equation with the general quadratic equation:
[tex]ax^2+bx+c=0[/tex]Our values for a, b, and c are:
[tex]\begin{gathered} a=1 \\ b=-18 \\ c=70 \end{gathered}[/tex]Step 3. Using the value of b, find the following expression:
[tex](\frac{b}{2})^2[/tex]The result is:
[tex](-\frac{18}{2})^2\longrightarrow(-9)^2[/tex]Step 4. Take the original equation
[tex]x^2-18x+70=0[/tex]Move the +70 as a -70 to the right-hand side:
[tex]x^2-18x=-70[/tex]And now add to both sides the expression found in step 3 for (b/2)^2:
[tex]x^2-18x+(-9)^2=-70+(-9)^2[/tex]Step 5. Factor the left-hand side of the equation as a perfect square binominal:
[tex]\begin{gathered} P\operatorname{erf}ect\text{ square binomial formula:} \\ (a\pm b)^2=a^2\pm2ab+b^2 \end{gathered}[/tex]Applying this to our expression:
[tex](x-9)^2=-70+(-9)^2[/tex]Step 6. Solve the operations on the right-hand side:
[tex]\begin{gathered} (x-9)^2=-70+81 \\ \downarrow\downarrow \\ \boxed{\mleft(x-9\mright)^2=11} \end{gathered}[/tex]The form is the equation is:
[tex]\boxed{(x-9)^2=11}[/tex]Step 7. To find the value or values of x, solve for x in the previous equation:
[tex]\begin{gathered} (x-9)^2=11 \\ \downarrow\downarrow \\ x-9^{}=\pm\sqrt[]{11} \\ \downarrow\downarrow \\ x^{}=\pm\sqrt[]{11}+9 \end{gathered}[/tex]Step 8. To find the two solutions we use the '+' and '-' signs separately:
[tex]\begin{gathered} x^{}=\sqrt[]{11}+9\longrightarrow x=3.3166+9=12.3166 \\ x^{}=-\sqrt[]{11}+9\longrightarrow x=-3.3166+9=5.6834 \end{gathered}[/tex]Rounding these values for x to the nearest hundredth (2 decimal places):
[tex]\begin{gathered} x=12.32 \\ x=5.68 \end{gathered}[/tex]Answer:
Form:
[tex]\boxed{(x-9)^2=11}[/tex]Solution:
[tex]x=12.32,5.68[/tex]Two linear functions are shown below. Compare each fuoction to answer the questions. Function 2: Function 1: -11 8 -7 13 3 Ng -3 18 Part A: What is the rate of change for Function 1? Part B: What is the rate of change for Function 2? Part C: Which function has the greater rate of change?
The rate of change of a linear functions is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]where (x1,y1) and (x2,y2) are points through the graph.
Function 1.
From the table we have that the functions passes through the points (-11,8) and (-7,13), pluggin the values in the formula above we have:
[tex]\begin{gathered} m=\frac{13-8}{-7-(-11)} \\ m=\frac{5}{11-7} \\ m=\frac{5}{4} \end{gathered}[/tex]Therefore the rate of change of functions 1 is 5/4
Function 2.
From the graph we notice that the functions passes through the points (-3,-4) and (1,-1), hence:
[tex]\begin{gathered} m=\frac{-1-(-4)}{1-(-3)} \\ m=\frac{-1+4}{1+3} \\ m=\frac{3}{4} \end{gathered}[/tex]Therefore the rate of change of function 2 is 3/4.
Comparing both rates of change we conclude that Function 1 has the greater change of rate.