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3 years ago

11

Guys pls help me.........

1 answer:

Nezavi [6.7K]3 years ago

7 0

What seems to be your problem? Are you okay? I hope everything is all right :)

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The maximum weight on a roller coaster is 400 pounds. If your friend weighs 120 pounds, what is the most that you can weigh so y

wolverine [178]

The maximum weight on a roller coaster is 400 pounds (Max = 400).
If your friend weighs 120 pounds (y = 120), what is the most you can weigh so you can ride together (x + y < 400).

x + 120 <_ 400.
y = 120

Subtract 120 from both sides.

x <_ 280.

You can weigh at most 280 pounds (x <_ 280).

I hope this helps!

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3 years ago

Cot theta is undefiend for theta

Elena-2011 [213]

Cot theta is 1 over tan theta.The only way for 1/tan theta to be undefined is for tan theta = 0. - answers

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Which answer shows a correct order to solve this story problem?

OLga [1]

The correct answer would be C :^)

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if you have been ask to offer your services wrapping gifts at the catholic womans league bazar. you have to wrap the following b

34kurt

Answer:

wowowowowowow cool

Step-by-step explanation:

the answer is easy nobody cares :)

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3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago