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alexandr1967 [171]
3 years ago
7

Why do equivalent ratios form a straight line when graphed? O The vertical distance between points is constant, and the horizont

al distance between points increases with each point. O The ratio of the change in y-values to the change in x-values is equivalent for any two points in a graph of equivalent ratios. O The horizontal distance between points is constant, and the vertical distance between points increases with each point. O The y-values of equivalent ratios increase at a constant rate, and the x-values decrease at a constant rate.​
Mathematics
2 answers:
hoa [83]3 years ago
7 0

Answer:

D-The y-values of equivalent ratios increase at a constant rate, and the x-values decrease at a constant rate.

Step-by-step explanation:

did the quiz:)

Bumek [7]3 years ago
5 0

hes right, its d :)

good luck on your test!

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The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
3 years ago
Find the slope of the line graphed below
igomit [66]

We are given two points: (-1, -1) and (1, -4). Slope is calculated as the change in y over the change in x, or rise over run.

The change in y is the difference of the two y coordinates (it doesn't matter the order): -1 - (-4) = 3

The change in x is the difference of the two x coordinates (this order depends on the order that you subtracted the y coordinates; they must be the same order): -1 - 1 = -2.

So, the slope is 3/-2

4 0
3 years ago
Read 2 more answers
Which espression is equivalent to 7^(-2)*7^(6)
valina [46]

Answer: it’s A and D

Step-by-step explanation:

8 0
3 years ago
And here is the other one.
LUCKY_DIMON [66]
Since BD bisects angle ABC, that means angle ABD and angle CBD are equal to each other. With that set up the equation to solve for x like this:

-4x+33 = 2x+81
-2x -2x
————————
-6x +33 = 81
-33 -33
————————
-6x = 48
————- (divide by -6)
-6

x = -8


Now substitute that to ABD
-4(-8) +33
32 +33
=65

here’s CBD
2(-8) + 81
-16 + 81
=65

Finally angle ABC will be double the amount of ABD or CBD so 65 times 2 is 130.

ANSWERS: (angles)
ABD and CBD: 65
ABC: 130

3 0
3 years ago
Find the y-intercept and the x-intercept of the line 2x-4y=10
kompoz [17]
Finding y intercept and x intercept is easy:

X intercept will be of the form (x,0) and y intercept will be of the form (0,y)

● If you put x=0 in the equation, you will get y-intercept.

● If you put y=0 in the equation, you will get x-intercept.
______________________________

Given equation: 2x - 4y = 10

◆ Put x = 0
2×0 - 4y = 10
=> -4y = 10
=> y = 10/(-4)
=> y = -5/2

Thus y intercept is (0, -5/2)

◆Put y = 0
2x - 4×0 = 10
=> 2x = 10
=> x = 10/2
=> x = 5

Thus the x intercept is (5,0)
4 0
3 years ago
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