For this exercise it is important to know that, in Dilations:
1. When the scale factor is greater than 1, the Image (the figure obtained after the transformation) is an enlargement.
2. When the scale factor is greater than 0 but less than 1, the Image is a reduction:
![0You know that the coordinates of the vertices of the polygon ABCD are:[tex]\begin{gathered} A(2,4) \\ B(-4,-8) \\ C(0,4) \\ D(12,-2) \end{gathered}](https://tex.z-dn.net/?f=0You%20know%20that%20the%20coordinates%20of%20the%20vertices%20of%20the%20polygon%20ABCD%20are%3A%5Btex%5D%5Cbegin%7Bgathered%7D%20A%282%2C4%29%20%5C%5C%20B%28-4%2C-8%29%20%5C%5C%20C%280%2C4%29%20%5C%5C%20D%2812%2C-2%29%20%5Cend%7Bgathered%7D)
And the scale factor is:
Since:
It is a reduction.
You can identify that the rule of this transformation is:
Answer:
B
300 ft
Perimeter of new triangle
In order to find how long the the kangaroo's jump was, we are going to find out what the value of x is when y=0, because y=0 represents the kangaroo on the ground.
0 = -0.03(x - 14)2 + 6 Original equation with 0 substituted for y
0.03(x - 14)2 = 6 Add 0.03(x - 14)2 to each side
(x - 14)2 = 200 Divide both sides by 0.03
x - 14 = +/- 10√2 Take the square root of each side
x = 14 +/- 10√2 Add 14 5o each side
Because 10√2 is approximately 14.14, 14 - 10√2 is approximately 0. This makes sense because the kangaroo has not left the ground yet at 0 ft. The correct answer is 14 + 10√2, which is approximately 28.14 ft.
For part a, we want to use the distance that is halfway between the zeroes of the jump. That is the point halfway between 14 - 10√2 and 14 + 10√2. In other words, we need to use the value 10√2 (approx. 14.14) for x and solve for y.
y = -0.03(10√2 - 14)2 + 6 Write original equation with 10√2 substituted for y
y = -0.03(14.14 - 14)2 + 6 (Using approximation)
y = -0.03(0.14)2 + 6 Substitute (14.14 - 14 = 0.14)
y = -0.03(0.0196) + 6 Square 0.14
y = -0.000588 + 6 Multiply -0.03 times 0.0196
y = 5.999412 Add -0.00588 and 6
y = 6 Round
I went through all of the this work to show that the maximum point of the parabola (the highest point in the kangaroo's jump) is actually the point where the value being squared is 0 (14 - 14). The y-value of the vertex is 6. That is also the highest point in the jump.
The answers are:
a. 6 ft
b. 28.14 ft
The opposite of. 0.5 or 1/2
