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3 years ago

Which of the following is equal to the expression below?

Mathematics

2 answers:

choli [55]3 years ago

7 0

Answer: OPTION C.

Step-by-step explanation:

By definition:

$a^{1/n}=\sqrt[n]{a}$

You must descompose the numbers 160 and 243 into their prime factor:

$160=2*2*2*2*2*5=2^{5}*5$

$243=3*3*3*3*3=3^{5}$

Simplify the expression, then you obtain the following result:

$\sqrt[5]{2^{5}*5*3^{5}}=2*3\sqrt[5]{5}=6\sqrt[5]{5}$

The answer is the option C.

AnnZ [28]3 years ago

7 0

<h2>Answer:</h2>

<u>The exact answer is option </u><u>C</u>

<h2>Step-by-step explanation:</h2>

Look at the attached picture for solution

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How many 3/4 cup servings are in 12 cups of cherries?

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Answer:

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Step-by-step explanation:

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Using unitary method to solve.

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Step-by-step explanation:

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(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a

Mazyrski [523]

The length of a curve $C$ given parametrically by $(x(t),y(t))$ over some domain $t\in[a,b]$ is

$\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

In this case,

$x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)$

$y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)$

So we have

$\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t$

and the arc length is

$\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt$

We have

$\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4$

where $n$ is any integer; this tells us $\sin(4t)\ge0$ on the interval $\left[0,\frac\pi4\right]$ and $\sin(4t)$ on $\left[\frac\pi4,1\right]$ . So the arc length is

$=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)$

$=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)$

$=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}$

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3 years ago