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Romashka [77]
3 years ago
14

Chase was given a box of assorted chocolates for his birthday. Each night, Chase treated himself to some chocolates. There were

originally 18 chocolates in the box and after 4 nights, there were 6 chocolates remaining in the box. Write an equation for C,C, in terms of t,t, representing the number of chocolates remaining in the box tt days after Chase's birthday.
Mathematics
2 answers:
masya89 [10]3 years ago
8 0

Answer:

Me scrolling to see if anybody needs help that I am really not confused on

Step-by-step explanation:

wel3 years ago
8 0

Answer:

C=-3t+18

Step-by-step explanation:

I just did it.

He ate 12 chocolates in total:18-12=6

So he ate for 4 nights which means he ate 3 chocolates every night

t represents the days which is how many chocolates he ate and 18 was how

many Chocolates were in total! Done u just put it in the equation. I hope this helps. <3

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Answer:

7\sqrt{3}

Step-by-step explanation:

Since this is a 30-60-90 triangle, we know that the sides have the following characteristic:

The side opposite to 30 degree angle: n

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3 years ago
3y/5 = 15 the value of y is
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Step-by-step explanation:

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3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
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where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
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Dvinal [7]

Answer:

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Step-by-step explanation:

because

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3 years ago
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nasty-shy [4]

Answer:

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Step-by-step explanation:

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