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CaHeK987 [17]
3 years ago
6

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

Mathematics
1 answer:
cricket20 [7]3 years ago
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.

4 0
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<br> 90<br> 80<br> 70<br> 60
cupoosta [38]

Answer:

75

Step-by-step explanation:

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

 (90 + 80 + 70 + 60) / 4 = 75

8 0
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 <span>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.
</span>
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
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