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lutik1710 [3]
3 years ago
5

Which measurement is most accurate to describe the amount that a teacup can hold?

Mathematics
2 answers:
Anni [7]3 years ago
5 0
ML milliliters  hope that helps

eduard3 years ago
4 0
Milliliters, they aren't the absolute smallest but they are still small since a teacup is small.. hope I helped:)
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Find the missing number in this function table.
kirill115 [55]

Answer:

25

Step-by-step explanation:

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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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The correct answer should be -3/2

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principal phillipi used a random sample of 20 students records to determine how far, in miles, students live from the school. Th
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To find the mean:

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