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Alexeev081 [22]
3 years ago
12

Factor completely.

Mathematics
2 answers:
nexus9112 [7]3 years ago
8 0
5x^2+10x-40=5(x^2+2x-8)=5(x^2+4x-2x-8)=5[x(x+4)-2(x+4)]=5(x-2)(x+4)

The Answer: B

:)
Burka [1]3 years ago
6 0
5x² + 10x - 40

Let's go through all of the answers & simplify them, then see if they are correct.

A) 5(x - 4)(x + 2)

Simplify.

5x² - 10x - 40

So, A) is INCORRECT.

B) 5(x - 2)(x + 4)

Simplify.

5x² + 10x - 40

So, B) is the correct answer.

However, let's also check C just in case :)

C) 5(x - 4)(x - 2)

Simplify.

5x² - 30x + 40

Therefore C) is incorrect.

So, your correct answer is B) 5(x - 2)(x + 4)

~Hope I helped!~

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For the following exercises, evaluate the function at the indicated values:
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Answer:

The evaluated function for the indicated values is given below.

The value of f(-3) is 20 .

The value of f(2) is 10 .

The value of f(-a) is $2|-3 a-1|$.

The value of -f(a) is $-2|3 a-1|$.

The value of f(a+h) is $2|3 a+3 h-1|$.

Step-by-step explanation:

A function $f(x)=2|3 x-1|$ is given.

It is required to evaluate the function at $f(-3) ; f(2) ; f(-a) ;-f(a) ; f(a+h)$.

To evaluate the function, substitute the indicated values in the given function to determine the output values and simplify the expression.

Step 1 of 5

The given function is $f(x)=2|3 x-1|$.

To evaluate the function at f(-3), substitute -3 in the given function f(x)=2|3 x-1|$\\ $f(-3)=2|3(-3)-1|$\\ $f(-3)=2|-9-1|$\\ $f(-3)=2|-10|$\\ $f(-3)=2(10)$\\ $f(-3)=20$

Step 2 of 5

To evaluate the function at $f(2)$, substitute 2 in the given function.

f(x)=2|3 x-1|$\\ $f(2)=2|3(2)-1|$\\ $f(2)=2|6-1|$\\ $f(2)=2(5)$\\ $f(2)=10$

Step 3 of 5

To evaluate the function at f(-a), substitute -a in the given function.

$$\begin{aligned}&f(x)=2|3 x-1| \\&f(-a)=2|3(-a)-1| \\&f(-a)=2|-3 a-1|\end{aligned}$$

Step 4 of 5

To evaluate the function at -f(a), substitute a in the given function.

f(x)=2|3 x-1|$\\ $-f(a)=-2|3(a)-1|$\\ $-f(a)=-2|3 a-1|$

Step 5 of 5

To evaluate the function at f(a+h), substitute a+h in the given function. $f(x)=2|3 x-1|$

$$\begin{aligned}&f(a+h)=2|3(a+h)-1| \\&f(a+h)=2|3 a+3 h-1|\end{aligned}$$

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