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2 years ago

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A square sandbox has sides of length 9 ft and a depth of 1 ft. If sand costs $5 / cubic foot, how much will it cost to fill the

sandbox with sand?

1 answer:

Leto [7]2 years ago

7 0

The cost spent on fill the box with sand is 405$.

According to the statement

we have given a square sandbox with length 9 ft and depth of 1 ft.

and the rupees of sand per cubic ft is $5

then

Now, we have to find the volume of sandbox with the given length and depth.

So, for this purpose we use Volume formula

V = (a)^3

V = 9*9*1

V = 81 cubic feet

And the cost of sand per cubic feet is $5.

then

The cost of sand for 81 cubic per feet is calculate by multiplication formula then

Total cost of sand = 81*5

Total cost of sand = 405$

So, The cost spent on fill the box with sand is 405$.

Learn more about VOLUME here brainly.com/question/11408596

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The lengths of two sides of a right triangle are 5 inches and 8 inches. What is the difference between the two possible

3241004551 [841]

Answer:

Option 2 - 3.2 inches.

Step-by-step explanation:

Given : The lengths of two sides of a right triangle are 5 inches and 8 inches.

To find : What is the difference between the two possible lengths of the third side of the triangle?

Solution :

According to question, it is a right angle triangle

Applying Pythagoras theorem,

$H^2=P^2+B^2$

Where, H is the hypotenuse the longer side of the triangle

P is the perpendicular

B is the base

Assume that H=8 inches and B = 5 inches

Substitute the value in the formula,

$8^2=P^2+5^2$

$64=P^2+25$

$P^2=64-25$

$P^2=39$

$P=\sqrt{39}$

$P=6.24$

Assume that P=8 inches and B = 5 inches

Substitute the value in the formula,

$H^2=8^2+5^2$

$H^2=64+25$

$H^2=89$

$H=\sqrt{89}$

$H=9.43$

Therefore, The possible length of the third side of the triangle is

$L=H-P$

$L=9.43-6.24$

$L=3.19$

Therefore, The difference between the two possible lengths of the third side of the triangle is 3.2 inches.

So, Option 2 is correct.

3 0

3 years ago

Read 2 more answers

How can you check your work when you are finding arc measures?

Murrr4er [49]

After you measured it you can check if the measurement is correct by knowing radius and arc length using formula,

$\alpha=\dfrac{l}{r}$ .

Hope this helps.

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3 years ago

How many different integers between $100$ and $500$ are multiples of either $6,$ $8,$ or both?

nirvana33 [79]

We need to find the number of integers between 100 and 500 that can be divided by 6, 8, or both. Now, to do this, we must as to how many are divisible by 6 and how many are multiples of 8.

The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have

$n = \frac{498-102}{6} + 1$
$n = 67$

We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.

$n = \frac{496-104}{8} + 1$
$n = 50$

That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.

$n = \frac{480-120}{24} + 1$
$n = 16$

Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.

Answer: 101

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Is chef is going to use a mixture of two brands of Italian dressing the first brand contains 9% vinegar and the second one conta

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Answer:

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2 years ago

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3 years ago