32% i believe because i added 4 an 2 together from the spinner an coin an got 6 the i divided 1 by 6 to get .16 and then doubled it because the precents before that doesnt matter if your looking for a specific combination
Answer:
The y-values of equivalent ratios increase at the same rate as their x-values. The vertical distance between points is constant, and the horizontal distance between points is constant. This forms a straight line.
Step-by-step explanation:
Given: Angle OQR and angle RQS form a linear pair and measure of angle OQR =
and measure of angle RQS= ![11x-65^{\circ}](https://tex.z-dn.net/?f=%2011x-65%5E%7B%5Ccirc%7D%20)
To find: The measure of angle RQS.
Solution:
Since Angle OQR and RQS forms linear pair.
So, the sum of their angles is 180 degrees.
![\angle OQR+\angle RQS=180^{\circ}](https://tex.z-dn.net/?f=%20%5Cangle%20OQR%2B%5Cangle%20RQS%3D180%5E%7B%5Ccirc%7D%20)
![5x+5^{\circ}+11x-65^{\circ}=180^{\circ}](https://tex.z-dn.net/?f=%205x%2B5%5E%7B%5Ccirc%7D%2B11x-65%5E%7B%5Ccirc%7D%3D180%5E%7B%5Ccirc%7D%20)
![16x-60^{\circ}=180^{\circ}](https://tex.z-dn.net/?f=%2016x-60%5E%7B%5Ccirc%7D%3D180%5E%7B%5Ccirc%7D%20)
![16x=180^{\circ}+60^{\circ}](https://tex.z-dn.net/?f=%2016x%3D180%5E%7B%5Ccirc%7D%2B60%5E%7B%5Ccirc%7D%20)
![16x=240^{\circ}](https://tex.z-dn.net/?f=%2016x%3D240%5E%7B%5Ccirc%7D%20)
![x=\frac{240^{\circ}}{16}](https://tex.z-dn.net/?f=%20x%3D%5Cfrac%7B240%5E%7B%5Ccirc%7D%7D%7B16%7D%20)
![x=15^{\circ}](https://tex.z-dn.net/?f=%20x%3D15%5E%7B%5Ccirc%7D%20)
Now, we will find the measure of angle RQS
![\angle RQS = 11x-65^{\circ}](https://tex.z-dn.net/?f=%20%5Cangle%20RQS%20%3D%2011x-65%5E%7B%5Ccirc%7D%20)
![\angle RQS = (11 \times 15^{\circ})-65^{\circ}](https://tex.z-dn.net/?f=%20%5Cangle%20RQS%20%3D%20%2811%20%5Ctimes%2015%5E%7B%5Ccirc%7D%29-65%5E%7B%5Ccirc%7D%20)
![\angle RQS = 100^{\circ}](https://tex.z-dn.net/?f=%20%5Cangle%20RQS%20%3D%20100%5E%7B%5Ccirc%7D%20)
Answer: its A. kn=NM
Step-by-step explanation:
Answer: The vertex is (-3, 2)
The equation y = -4(x + 3)^2 + 2 can be written as y = -4(x - (-3))^2 + 2
I rewrote x+3 as x - (-3)
Compare y = -4(x - (-3))^2 + 2 with the general template y = a(x-h)^2 + k
Note how
a = -4
h = -3
k = 2
So the vertex is (h,k) = (-3, 2)