Answer:
THANKS BRO
Step-by-step explanation:
Remember that a parallel line has the same slope...
So we already know the slope of the line containing point C is -4.
Using point slope equation, we can do...
y-y₁=m(x-x₁)
y-7= -4(x-4)
y-7- -4x+16
*Add seven to both sides...* y=-4x+23
There is you answer... hope this helps!
Answer:
$111
Step-by-step explanation:
37x3 = 111
111/3 =37
![f(x)=5x^2-5x-6 \\ \quad \\ \begin{cases} f(\boxed{x+h})=5(\boxed{x+h})^2-5(\boxed{x+h})-6 \end{cases}\qquad thus \\ \quad \\ \cfrac{f(x+h)-f(x)}{h}\qquad \textit{will be then} \\ \quad \\ \cfrac{[5({x+h})^2-5({x+h})-6]\quad -\quad [5x^2-5x-6]}{h} \\ \quad \\ \cfrac{[5(x^2+2xh+h^2)-5(x+h)-6]-[5x^2-5x-6]}{h} ](https://tex.z-dn.net/?f=f%28x%29%3D5x%5E2-5x-6%0A%5C%5C%20%5Cquad%20%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0Af%28%5Cboxed%7Bx%2Bh%7D%29%3D5%28%5Cboxed%7Bx%2Bh%7D%29%5E2-5%28%5Cboxed%7Bx%2Bh%7D%29-6%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ccfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%5Cqquad%20%5Ctextit%7Bwill%20be%20then%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ccfrac%7B%5B5%28%7Bx%2Bh%7D%29%5E2-5%28%7Bx%2Bh%7D%29-6%5D%5Cquad%20-%5Cquad%20%5B5x%5E2-5x-6%5D%7D%7Bh%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ccfrac%7B%5B5%28x%5E2%2B2xh%2Bh%5E2%29-5%28x%2Bh%29-6%5D-%5B5x%5E2-5x-6%5D%7D%7Bh%7D%0A)
and surely, you'd know what that is
