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3 years ago

Which expression is equivalent to 3/56x^7y^5 , if x ≠0 and y≠0

Mathematics

1 answer:

Mariulka [41]3 years ago

3 0

Answer:

$2x^2y\sqrt[3]{7xy^2}$

Step-by-step explanation:

Given

$\sqrt[3]{56x^7y^5}$

Required

Solve

Expand

$\sqrt[3]{8*7*x^6*x*y^3*y^2}$

Rewrite as:

$\sqrt[3]{8*x^6*y^3*7*x*y^2}$

Split

$\sqrt[3]{8*x^6*y^3} *\sqrt[3]{7*x*y^2}$

Express 8 as 2^3

$\sqrt[3]{2^3*x^6*y^3} *\sqrt[3]{7*x*y^2}$

Apply law of indices

$2^{(3/3)}*x^{(6/3)}*y^{(3/3)} *\sqrt[3]{7*x*y^2}$

$2*x^2*y *\sqrt[3]{7*x*y^2}$

$2x^2y *\sqrt[3]{7xy^2}$

$2x^2y\sqrt[3]{7xy^2}$

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3 years ago

Nadia is a stockbroker. She earns 13% commission each week. Last week, she sold $4,500 worth of stocks. How much did she make

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3 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

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Answer:

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Step-by-step explanation:

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