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3 years ago

If (x – 5) is a factor of f(x), which of the following must be true?

Mathematics

2 answers:

Margaret [11]3 years ago

8 0

I think no because what is f and how we know if its the same number

or maybe ...

Elanso [62]3 years ago

5 0

Answer:

A root of f(x) is x = 5.

Step-by-step explanation:

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Mr. Mueller is hungry, but he can't leave the fire station to get food. He's

marin [14]

Answer:

$3.97x + $10 < (or equal to) $25

Step-by-step explanation:

8 0

2 years ago

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Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

3 years ago

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Sales at a book store increased from 5000 to 10000. This year sales dicreased to 5000 from 10000.

erastovalidia [21]

It's basically 5,000*2=10,000
and 10,000/2=5,000. your answer would be either they multiplied 5,000 by 2 and 10,000 divided by 2, they increased the sales by 5,000

6 0

3 years ago

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Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago

PLEASE HELP!!!!!!

expeople1 [14]

The answer is forty five.

6 0

3 years ago

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