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Verdich [7]
3 years ago
8

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

than or equal to 0
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

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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puteri [66]

Answer:

<em></em>

<em>The team must win 2 games to have a win : loss ratio o 2.</em>

Step-by-step explanation:

  • <u>Ratio win : loss, r</u>:

       r = total wins / total losses

       

  • <u>Total wins, Tw</u>:

        Call n the number of new wins:

        Tw =number of wins until so far + number of new win = 6 + n

  • <u>Total losses, Tl</u>:

        Tl = number of losses so far + number of new losses = 4 + 0 = 4

  • <u>Target ratio</u>:

        r = 2 ⇒ Tw / Tl = (6 + n) / 4 = 2

Solve for n:

  • (6 + n ) = 4 × 2
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  • n = 8 - 6
  • n = 2

Hence,<em> the team must win 6 more games.</em>

  • <u>Verification</u>: (6 + 2 ) / 4 = 8 / 4 = 2, which is the target ratio.

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Step-by-step explanation:

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Step-by-step explanation:

I understand this that we need to find a point B for each of a) to f).

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the slope 2/3 means that when going to B x has to increase by 3, and y has to increase by 2.

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-2/3 means that either x or y have to decrease and the other to increase.

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1 year ago
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Olegator [25]
Answer: Choice C) 

--------------------------------------

mean = xbar = (2400+1750+1900+2500+2250+2100)/6
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Subtract the data values from the mean to get

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2500-2150 = 350
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The differences are: 250, -400, -250, 350, 100, -50 

Then you square those values and add up the squares

(250)^2 +  (-400)^2 +  (-250)^2 +  (350)^2 +  (100)^2 +  (-50)^2 = 420,000
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