Answer:
Step-by-step explanation:
LOL The graph doesn’t match the y intercept :)
Anyway
If we have point (0,-3) we have a quadratic of
y=ax^2+bx-3 we are given points (-1,0) and (2,0) so
a-b-3=0 and 4a+2b-3=0
4a+2b-3+2(a-b-3)=0
4a+2b-3+2a-2b-6=0
6a-9=0
6a=9
a=1.5, since a-b=3
1.5-b=3
b=-1.5
y=1.5x^2-1.5x-3
<h2><u>Given</u><u>:</u><u>-</u></h2>
<h2><u>Solution</u><u>:</u><u>-</u></h2>
Move ABCD 3 units to the left and 2 units up. (x-3, y+2)
Step-by-step explanation:
Given
f(x) = 8x - 9
Then
For a.
f(3y - 1 ) = 8(3y - 1 ) - 9
= 24y - 8 - 9
= 24y - 17
Now for b.
f(x) = - 23
or, 8x - 9 = - 23
8x = - 23 + 9
8x = - 14
x = - 14 / 8
x = - 7 / 4
Hope it will help :)
Answer:
Step-by-step explanation:
Let, 
= y
sin(y) = 
---------(1)
cos(y) = 
= 
= 
Therefore, from equation (1),
Or ![\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7Bx%7D%7B6%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7Bx%5E2%7D%7B36%7D%7D%7D)
At x = 4,
![\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7B4%7D%7B6%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7B4%5E2%7D%7B36%7D%7D%7D)
![\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7B2%7D%7B3%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7B16%7D%7B36%7D%7D%7D)
