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nekit [7.7K]
4 years ago
9

Which expression is equivalent to StartFraction (5 a b) cubed Over 30 a Superscript negative 6 Baseline b Superscript negative 7

Baseline EndFraction? Assume a not-equals 0, b not-equals 0.
StartFraction a Superscript 7 Baseline b Superscript 10 Baseline Over 6 EndFraction
StartFraction 125 a Superscript 18 Baseline b Superscript 21 Baseline Over 30 EndFraction
StartFraction 25 a cubed b Superscript 4 Baseline Over 6 EndFraction
StartFraction 25 a Superscript 9 Baseline b Superscript 10 Baseline Over 6 EndFraction
Mathematics
1 answer:
m_a_m_a [10]4 years ago
7 0

Answer:

Option D.

Step-by-step explanation:

The given expression is

\dfrac{(5ab)^3}{30a^{-6}b^{-7}}

We need to find the expression, which is equivalent to the given expression.

The given expression can be rewritten as

\dfrac{5^3a^3b^3}{30a^{-6}b^{-7}}     [\because (ab)^m=a^mb^m]

\dfrac{125}{30}\times \dfrac{a^3}{a^{-6}}\times \dfrac{b^3}{b^{-7}}

\dfrac{25}{6}\times a^{3-(-6)}\times b^{3-(-7)}    [\because \dfrac{a^m}{a^n}=a^{m-n}]

\dfrac{25}{6}\times a^{3+6}\times b^{3+7}

\dfrac{25a^9b^{10}}{6}

Therefore, the correct option is D.

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Y=x.arctan(x)^1/2. find dy/dx. pls show steps​
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y=x(\arctan x)^{1/2}

Use the product rule first:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

Use the chain rule to compute the derivative of (\arctan x)^{1/2}. Let z=(\arctan x)^{1/2} and take u=\arctan x, so that by the chain rule

\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}

\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}

\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}

So we have

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}

You can rewrite this a bit by factoring (\arctan x)^{-1/2}, just to make it look neater:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)

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Answer:

Mean: 4.2

Median: 4

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Answer:

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Step-by-step explanation:

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Answer:

2x10+6=16

that's the equation

the expression doesn't have an answer so it would look like this:

2x10+6

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Willie scored 20% higher on his second math test than his first. If his first score was 82, what was his math test, rounded to t
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20% = 0.02
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