All categories

3 years ago

Collins is working.......

Mathematics

1 answer:

lesya [120]3 years ago

4 0

Let n represent the amount Colin earned on Sunday.

On Sat. he earned n/2; on Sun. he earned n; and on Friday he earned (1/2)(n/2).

Then n/2 + n + n/4 = $70

Mult. all terms by 4 to eliminate fractions:

2n + 4n + n = $280

7n = $280 => n = $40

Colin earned n/2, or $20, on Saturday; n, or $40, on Sunday; and n/4, or $10, on Friday.

Note that $20 and $40 and $10 add up to $70, as they must.

You might be interested in

85 is greater than the difference of x and63

miss Akunina [59]

85 > x-63

This is how you would arrange the problem,

isolate the X by adding 63 to 85 to get 148,
Flip the inequality to <

resulting in X<148.

8 0

3 years ago

How do you find an explicit formula?

bazaltina [42]

If you can find an explicit formula for a sequence, you will be able to quickly and easily find any term in the sequence simply by replacing n with the number of the term you seek. An explicit formula designates the nth term of the sequence, as an expression of n (where n = the term's location).

6 0

3 years ago

One kilogram is 2.2 pounds. Complete the tables. What is the interpretation

valentinak56 [21]

You would multiply each kilogram amount by 2.2 to get the point equivalent.

3 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

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

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

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

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago

Is 2 and 4.5 equivalent

o-na [289]

2 and 4.5 is not equivilant since 2 and 4.5 cannot go together (the 0.5 does not help and so it does not make it equivalent.)

Hope this helped!

Nate

3 0

3 years ago

Read 2 more answers