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3 years ago

A box has a volume given by the trinomial x^3+4x^2-5x. What are possible dimensions of the box? Use factoring.

1 answer:

6 0

Answer: The correct answer is Choice B.

To factor this expression, we should first factor out the variable x.

We will have:
x(x^2 + 4x - 5)

Now, all we have to do is factor x^2 + 4x - 5.

The factors of this are (x + 5)(x - 1).

If you put all of the factors together, you have: x(x + 5)(x - 1) or Choice B.

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Hannah is giving her sister a softball for her birthday, and wants to put it in a cubic gift box to preserve the surprise. She w

Inga [223]

Answer:

B. E(s) = s^3 - 29.

Step-by-step explanation:

The amount of empty space $E(s)$ inside the cubic box is equal to the volume of the box minus the volume of the softball.

If the side length of the cubic box is $s$ , and the volume of the softball is $29in^3$ , then the amount of empty space $E(s)$ inside is

$E(s) = s^3-29$ ,

where $s^3$ is the volume of the cubic box.

Since it matches the equation we got above, choice B stands correct.

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3 years ago

The two polygons are similar.Find the length of x

alexandr1967 [171]

Answer:

x = 9.2

Step-by-step explanation:

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3 years ago

What is the mean score Score 90 80 70 60

cupoosta [38]

Answer:

Step-by-step explanation:

Four scores.... add them together and divide by four

(90 + 80 + 70 + 60) / 4 = 75

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2 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

7 0

3 years ago

Solve the variables

MrRissso [65]

Answer:

x=9

y=15

Step-by-step explanation:

Its a parallel gram, so the sides would be the same as the parallel side,

3 0

3 years ago

Read 2 more answers