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Sav [38]
3 years ago
6

A total of 430 tickets were sold for the school play . They were either adult tickets or student tickets . There were 70 fewer s

tudent tickets sold than adult tickets .How many adult tickets were sold?
Mathematics
1 answer:
Maurinko [17]3 years ago
6 0
So first we assume that half is adult and have is student.

So
215 adult  215 student

Now since it said they were 70 less student tickets than adult tickets we subtract 70.
215 - 70 = 145

Now we add the 70 we subtracted from student tickets to the adult tickets.
215 + 70 = 285

So there were 
285 adult tickets sold and 145 student tickets sold

I hope this helped. Have a great day!
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Steve and Abby purchased a set of vases to place on a 12 foot long mantel above their fireplace. They want to place one vase 1/4
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Answer: One vase should be placed 3 feet from the end of the mantel and the other vase should be placed 9 feet from the same end.

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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;

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Solve exponents

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From laws of indices,

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\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}

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

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\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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