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lisabon 2012 [21]
3 years ago
6

Find the volume of this right rectangular prism. [Type your answer as a number.]

Mathematics
1 answer:
irakobra [83]3 years ago
3 0
There is no picture
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An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situati
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Answer:

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An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situation using money as the context. Then, evaluate the expression and interpret the result in terms of the situation. Here's an example.

The following situation is modeled by the expression -24.50 + 100.00: Jonathan overdrew his checking account by $24.50. He then transferred $100.00 from his savings account to his checking account. His new checking account total is $75.50.

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2) $20 - $18.04

3) 7($1.45)

4) 42.08 / 4

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What is the volume of a square pyramid with base edges of 60 inches and a height of 10 inches?
vichka [17]

Answer:

The volume is 12000 in^3

Step-by-step explanation:

From the question ,

L= 60\:in. and H=10\:in.

The volume of a pyramid is given as

V=\frac{1}{3} \times area\:of\:the\:base \times height

From the question, the base of the pyramid is a square.

Hence base area is given by L^{2}

This implies that the volume of the pyramid is

=\frac{1}{3} \times L^{2}\times H

=\frac{1}{3}\times 60^{2} \times 10

=\frac{36000}{3}

=12000 in^3

6 0
3 years ago
NO LINKS!!
dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

----------------------

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

4 0
2 years ago
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