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3 years ago

If 16 counters is 4/9 what is the ONE?

1 answer:

5 0

(4/9)/16 = 4/(9*16) = 4/(4*4*9) = 1/(4*9) = 1/36

One counter is $\frac{1}{36}$ <span> if 16 of them are </span> $\frac{4}{9}$

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g Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 7yi + xzj + (

Sati [7]

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

where $S$ is any oriented surface with boundary $C$ . We have

$\vec F(x,y,z)=7y\,\vec\imath+xz\,\vec\jmath+(x+y)\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-7)\,\vec k$

Take $S$ to be the ellipse that lies in the plane $z=y+9$ with boundary on the cylinder $x^2+y^2=1$ . Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+9)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

Then we have

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\big((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+2)\,\vec k\big)\cdot\big(-u\,\vec\jmath+u\,\vec k\big)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(3u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{3\pi}$

5 0

3 years ago

What is an approximate average rate of change of the graph from x = 2 to x = 5, and what does this rate represent? (4 points)

son4ous [18]

Answer:

Step-by-step explanation:

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3 years ago

Suppose you have a total of $2.25 in dimes and nickels and you have twice as many dimes as nickels. How much of each coin do I h

andreyandreev [35.5K]

Answer:

12 dimes and 21 nickles

Step-by-step explanation:

4 0

3 years ago

Given that cos θ = –12∕13 and θ is in quadrant III, find csc θ.

Trava [24]

Answer:

Step-by-step explanation:

We are given:

$\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}$

Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):

$o=\sqrt{13^2-12^2}=\sqrt{25}=5$

So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.

And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.

Cosecant is given by the hypotenuse over the opposite side. Thus:

$\displaystyle \csc(\theta)=\frac{13}{5}$

Since θ is in QIII, cosecant must be negative:

$\displaystyle \csc(\theta)=-\frac{13}{5}$

Our answer is A.

7 0

3 years ago

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There are 60 new houses being built in a neighborhood. Last month 1/3 of them were sold. This month 1/5 of the remaining houses

tatuchka [14]

Answer:

Step-by-step explanation:

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3 years ago