3 years ago

Eathan correctly answers 80% of the total questons

1 answer:

3 0

Answer:

he got 20% wrong

he got a B

Step-by-step explanation:

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<span>Least to greatest
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USPshnik [31]

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath+t\,\vec k$

$\implies\vec r'(t)=-2\sin t\,\vec\imath+2\cos t\,\vec\jmath+\vec k$

$\implies\|\vec r'(t)\|=\sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}=\sqrt5$

Then the length of the arc is

$\displaystyle\int_0^{2\pi}\|\vec r'(t)\|\,\mathrm dt=\sqrt5\int_0^{2\pi}\mathrm dt=2\sqrt5\,\pi$

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Juli2301 [7.4K]

Answer: p=28 inches

Step-by-step explanation:

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