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Nady [450]
2 years ago
10

Compare A and B in three​ ways, where A=1.84 million is the 2012 daily circulation of newspaper X and B=2.26 million is the 2012

daily circulation of newspaper Y.
a. Find the ratio of A to B.
b. Find the ratio of B to A.
c. Complete the​ sentence: A is​ ____ percent of B.
Mathematics
2 answers:
gavmur [86]2 years ago
6 0
C Complete the sentence a is _____ percent of b
lions [1.4K]2 years ago
6 0

Step-by-step explanation:

A=1.84 million is the 2012 daily circulation of newspaper X

B=2.26 million is the 2012 daily circulation of newspaper Y

a) The ratio of A to B

\frac{A}{B}=\frac{1.84 million}{2.26 million}=\frac{92}{113}

b) The ratio of B to A

\frac{B}{A}=\frac{2.26 million}{1.84million}=\frac{113}{92}

c)  A is​ <em><u>81.42</u></em> percent of B.

A=1.84 million

B=2.26 million

\frac{1.84 million}{2.26 million}\times 100=81.42\%

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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

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From laws of indices,

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

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

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