$\bf \textit{mean value theorem}\\\\
f'(c)=\cfrac{f(b)-f(a)}{b-a}\qquad 
\begin{cases}
a=-4\\
b=0
\end{cases}\implies f'(c)=\cfrac{f(0)-f(-4)}{0-(-4)}
\\\\\\
f'(c)=\cfrac{9-5}{0+4}\implies f'(c)=\cfrac{4}{4}\implies f'(c)=1$

4 0

3 years ago

Work out m and c for the line: y = x + 4

OLEGan [10]

Answer:

Step-by-step explanation:

6 0

3 years ago

The expression 2y+3y2−5y+y2simplifies to 4y2−3y <br> true or false

Mumz [18]

Answer:

False?

Step-by-step explanation:

I'm pretty sure it's false because wouldn't it simplify to 5y? Sorry if I'm wrong. :)

8 0

3 years ago

If f(x) = 5x − 1 and g(x) = 2x2 + 1, what is the value of (f × g)(-3)?

Softa [21]

f(x) = 5x − 1 and g(x) = 2x^2 + 1

(f × g)(x) = (5x − 1)(2x^2 + 1)

(f × g)(x) = 10x^3 - 2x^2 + 5x - 1

Substitute x = - 3

(f × g)(-3) = 10(-3)^3 - 2(-3)^2 + 5(-3) - 1

(f × g)(-3) = 10(-27) - 2(9) -15 - 1

(f × g)(-3) =-270 - 18 - 16

(f × g)(-3) = -236

Answer

- 236

6 0

3 years ago