3 3/4 times -2/5 A. - 1/2 B. -2 C.2 D. 1/2

If I have the height and volume of a cylinder, how do I find the radius? or diameter?

Anna71 [15]

If you have the volume, that's a pretty good giveaway that you ought to use
the formula for the volume of a cylinder, since it involves all the other things ...
the things you're given, and the things you need to find. If you know the formula
for the volume of a cylinder, you can whip out the answer in a second. If you don't
know the formula, then you are toast.

Volume of a cylinder = (pi) (radius)² (length or height)

Divide each side of the equation by (pi) x (height) and you have

(Radius)² = Volume / (pi x height)

You HAVE volume and height, so you can use this equation to calculate
the radius. Once you have that, if you need the diameter, it's just double
the length of the radius.

That's how.

6 0

3 years ago

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Evaluate the integral below, where C is the curver(t) = ‹sin(t),cos(t), sin(2t)›, 0 ≤ t ≤2π. (Hint: Observe that C lies on the s

nordsb [41]

We can compute the integral directly: we have

$\begin{cases}x(t)=\sin t\\y(t)=\cos t\\z(t)=\sin(2t)\end{cases}\implies\begin{cases}\mathrm dx=\cos t\,\mathrm dt\\\mathrm dy=-\sin t\,\mathrm dt\\\mathrm dz=2\cos(2t)\,\mathrm dt\end{cases}$

Then the integral is

$\displaystyle\int_C(y+9\sin x)\,\mathrm dx+(z^2+4\cos y)\,\mathrm dy+x^3\,\mathrm dz$

$=\displaystyle\int_0^{2\pi}\bigg((\cos t+9\sin(\sin t))\cos t-(\sin^2(2t)+4\cos(\cos t))\sin t+\sin^3t\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\cos^2t+9\cos t\sin(\sin t)-\sin^2(2t)\sin t-4\sin t\cos(\cos t)+\sin^3t\,\mathrm dt$

You could also take advantage of Stokes' theorem, which says the line integral of a vector field $\vec F$ along a closed curve $C$ is equal to the surface integral of the curl of $\vec F$ over any surface $S$ that has $C$ as its boundary.

In this case, the underlying field is

$\vec F(x,y,z)=\langle y+9\sin x,z^2+4\cos y,x^3\rangle$

which has curl

$\mathrm{curl}\vec F(x,y,z)=-\langle2z,3x^2,1\rangle$

We can parameterize $S$ by

$\vec s(u,v)=\langle u\cos v,u\sin v,2u^2\cos v\sin v\rangle=\langle u\cos v,u\sin v,u^2\sin(2v)\rangle$

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

Note that when viewed from above, $C$ has negative orientation (a particle traveling on this path moves in a clockwise direction). Take the normal vector to $S$ to be pointing downward, given by

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle2u^2\sin v,2u^2\cos v,-u\rangle$

Then the integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S\mathrm{curl}\vec F(x,y,z)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1-\langle2u^2\sin(2v),3u^2\cos^2v,1\rangle\cdot\langle2u^2\sin v,2u^2\cos v,-u\rangle\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^14u^4\sin(2v)\sin v+6u^4\cos^3v-u\,\mathrm du\,\mathrm dv$

Both integrals are kind of tedious to compute, but personally I prefer the latter method. Either way, you end up with a value of $\boxed\pi$ .

4 0

3 years ago

two runners are saving money to attend a marrathon. the first runner has $112 in savings, received a $45 gift from a friend, and

Arturiano [62]

112+45+25m=50+60m
Both sides should be qual to each other once they both hit a certain number of months. I hope this helps!

5 0

3 years ago

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How can you represent real world situations using positive nubers, negative nubers, and zero?

Natasha2012 [34]

Answer:Maybe ffor the positive numbers you could use like 1,2,3,4,5,6,7,8, and 9. For the negitive number could be like -1,-2,-3,-4,-5. And zero could repersent the middle. Get it or no?

Step-by-step explanation:

6 0

3 years ago

Use distributive property to model the product on the grid. .record the product 4*12=48

Darina [25.2K]

(2+2) * (6*2) = 4*12 = 48

4 0

3 years ago