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3 years ago

Help me I don’t get this.

Mathematics

1 answer:

Marizza181 [45]3 years ago

5 0

Answer:

i think its 105

Step-by-step explanation:

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The volume of a rectangular prism is represented by the function x3 + 11x2 + 20x – 32. The width of the box is x – 1 while the h

Oksi-84 [34.3K]

we know that

the volume of a rectangular prism is equal to

$V=L*W*H$

where

L is the length of the box

W is the width of the box

H is is the height of the box

in this problem we have

$V=x^{3\ }+11x^2+20x-32$

$W=x-1$

$H=x+8$

$L=?$

Find the length of the box

Using a graph tool-------> we will determine the roots of the equation of volume

see the attached figure

the roots are

$x=-8\\x=-4\\ x=1$

$x^{3\ }+11x^2+20x-32=(x+8)*(x+4)*(x-1)$

therefore

the answer is

the length of the box is equal to $(x+4)$

8 0

3 years ago

Read 2 more answers

5(3a-1)-2(3a-2)=3(a+2)+v

STatiana [176]

Answer:

The answer is option C. 6a-7

Step-by-step explanation:

Given that

5(3a-1)-2(3a-2)=3(a+2)+v

Solve for v

∴ v = 5(3a-1)-2(3a-2) - 3(a+2)

∴ v = 15a - 5 - 6a + 4 - 3a - 6

∴ v = 15a - 6a - 3a - 5 + 4 - 6

∴ v = 6a - 7

So the answer is option C. 6a-7

5 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago

7y - 4.5 - 6y = HELP ME IM SO CONFUSED

kotegsom [21]

Answer:

y=4.5

Step-by-step explanation:

8 0

3 years ago

There are 4 consecutive odd integers that sum to 120.

allsm [11]

Answer:

27, 29, 31, 33

Step-by-step explanation:

Let x represent the smallest odd integer then...

x+2 represents the second

x+4 represents the third

x+6 represents the fourth

We can use this to set up an equation:

x+x+2+x+4+x+6=120

Combine like terms

4x+12=120

Subtract 12 from both sides

4x=108

Divide both sides by 2

x=27

The integers are 27, 29, 31, 33

4 0

3 years ago