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2 years ago

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Mr. Bates and Mr. Z want to refinish the community center gym floor. The rectangular floor is 15.2 m wide and 25.6 long. If each

gallon of polyurethane covers about 180 square meters, how many gallons should Mr. Bates and Mr. Z buy? *

1 answer:

Zinaida [17]2 years ago

3 0

Answer:

3 gallons

Step-by-step explanation:

area = 15.2 x 25.6 = 389.12 m²

389.12/180 = 2.16

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What is the answer to<br> 3/8+ 5/6+1/2=

qaws [65]

Answer:

Fraction form: $\frac{41}{24}$

Decimal form: 1.71

Mixed number form: 1 $\frac{17}{24}$

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3 years ago

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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

Find the 8th term of the geometric sequence 7,28,112,...

oksian1 [2.3K]

Answer:

114,688

Step-by-step explanation:

the previous term is multiplied by 4 to get the next term.if you continue with this sequence you would eventually get 114 688

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2 years ago

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To save money, you put $200 in your bank account each week. After saving for 4 weeks, you have $1,700 dollars in your account. W

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Answer: y - 1700 = 200(x-4)

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3 years ago

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Rebecca went swimming yesterday. After a while she had covered one fifth of her intended distance. After swimming six more lengt

alukav5142 [94]

Answer:

Option E.

Step-by-step explanation:

Let the intended length of Rebecca's swimming is = x units

and we assume the length of the pool = l units

Now it is given in the question that " She covers one fifth of her intended distance "

That means distance covered = $\frac{x}{5}$

" After swimming six more lengths of the pool she had covered one quarter of her intended distance"

So $\frac{x}{5}+6(l)=\frac{x}{4}$

$6l=\frac{x}{4}-\frac{x}{5}$

$6l=\frac{x}{20}$

x = 20×(6l)

x = 120l

Therefore, Rebecca has to complete 120 lengths of the pool.

Option E is the answer.

4 0

3 years ago