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2 years ago

Marco had 21 party favors. He put them into 7 bags with the same

1 answer:

3 0

ANSWER:
3

EXPLANTION:
21/7 is 3

FUN FACT:
People have been celebrating birthdays since 100 AD.

I hoped this helped and have a wonderful day :)

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Wats the answer to the Lucy question

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I am sorry I have no idea what are you talking about

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3 years ago

A store sells a note book for $3 each an does not charge

Law Incorporation [45]

So what your question

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3 years ago

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Heyy! i’ll give brainliest please help.

san4es73 [151]

Answer:

The answer is B.

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3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

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2 years ago

What two numbers multiply to be -12 and add to be 1

stiks02 [169]

It’s 4 and negative 3

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2 years ago

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