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3 years ago

5

A brand of uncooked spaghetti comes in a box that is a rectangular prism with a length of 9 inches, a width of 2 inches, and a h

eight of 1 1/2 inches. What is the surface area of the box?

1 answer:

MrMuchimi3 years ago

7 0

<u>Answer:</u>

<h2>SA = 157 in²</h2>

<u>Steps:</u>

SA = 2LW + 2WH + 2LH

SA = 2(9×2) + 2(2×11/2) + 2(9×11/2)

SA = 2×18 + 2×22/2 + 2×99/2

SA = 36 + 22 + 99

SA = 157 in²

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CaHeK987 [17]

Answer:

No Solution

Step-by-step explanation:

| −2n | + 10 = −50

Subtract 10 on both sides

|-2n|+10-10=-50-10

Simplify

|-2n=|=-60

No Solution

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3 years ago

How do I decease 7 by 3%

FromTheMoon [43]

Answer:

6.79

Step-by-step explanation:

First of all, you will have to know how much is 3% of 7 in order to decrease it by that amount.

7 x 0.03 = 0.21

Now that you know the amount, you can subtract it from seven and therefore you have decreased 7 by 3%.

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2 years ago

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There are 9 computers and 72 students. what is the unit rate of students to computers?

stepan [7]

8 Is the unit rate. For 9/27

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3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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3 years ago

What is the ratio of odd numbers to even numbers in the<br> list below?<br> 9 2 4 37 1 6 8 14

PolarNik [594]

Ratio of odd numbers:even numbers
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Total odd numbers- 3
Even numbers are - 2, 4, 6, 8, 14
Total even numbers- 5

odd numbers:even numbers
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1 year ago

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