These tables represent a quadratic function with a vertex at (0, -1). What is the average rate of change for the interval from x= 9 to x= 10?
y
Interval
х
0
1
2.
Average rate
of change
1-2
1-2
17-2
- 7
3-2
9
- 11
3
0 to 1
1 to 2
2 to 3
3 to 4
4 to 5
5 to 6
-10
- 17
-26
-37
4
5
6
A. - 2
B.-82
C101
D.-19

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

angle efh= angle egh

it should be a.

hope it

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder