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3 years ago

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A formula for 8 bottles of window cleaner calls for 6 cups of rubbing alcohol, 2.25 gallons of water, and 1.5 cups of ammonia. H

ow many quarts of rubbing alcohol would the formula call for if a factory made 1,280 bottles for stores to sell to customers?

2 answers:

lawyer [7]3 years ago

8 0

Answer:

240 quarts

Step-by-step explanation:

If 8 bottles of window cleaner calls for 6 cups of rubbing alcohol and we want to know how many quarts of rubbing alcohol would the formula call for 1,280 bottles, the next proportion must be satisfied:

8/1280 = 6/x

x = 6*1280/8

x = 960 cups

We know that 1 cup = 0.25 quarts, then, the conversion of cup to quart is:

960*0.25 = 240 quarts

tamaranim1 [39]3 years ago

7 0

240 quarts----------

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Answer:

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Step-by-step explanation:

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2 years ago

Zen deposited some money in a savings account. The graph below shows the value of Zen's investment y, in dollars, after x months

kifflom [539]

The y-intercept of 1000 in the graph is at the point where x is equal to 0.

At $x=0$ , it means no month have passed. It is the initial point where Zen deposited the money.

What is the amount (y axis)? 1000!

So y represents the initial deposit of Zen, which is 1000.

ANSWER: Answer choice D (Amount of money Zen deposited in the savings account)

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3 years ago

Read 2 more answers

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

Can u help with this one u know who u are

tatiyna

It's the one on the top righr, -2/9 w + 2/15.

8 0

3 years ago

Read 2 more answers

After 200 feet of drilling on the first well, a soil test is taken. The probabilities of finding the particular type of soil ide

dimaraw [331]

Answer:

The revised probabilities are;

The probability of finding soil with oil = 0.8

The probability of finding soil with good oil = 0.16

The probability of finding medium quality oil = 0.64

Step-by-step explanation:

The given probability of finding soil with high quality oil, P(QO) = 0.20

The probability of finding soil with medium-quality oil, P(OM) = 0.80

The probability of finding soil with no oil, P(ON) = 0.2

Therefore, given that the probability of finding soil with no oil = 0.2, we have;

The probability of finding soil with oil, P(OP) = 1 - the probability of finding soil with no oil

P(OP) = 1 - 0.2 = 0.8

Which gives;

The probability, P(FG) of finding soil with oil and that the oil is good is given as follows;

P(FG) = P(QO) × P(OP) = 0.2 × 0.8 = 0.16

The probability of finding good oil = 0.16

Similarly;

The probability of finding medium quality oil P(FM) = P(OM) × P(OP) = 0.8 × 0.8 = 0.64

Which gives the revised probability as follows;

The probability of finding soil with oil, P(OP) = 0.8

The probability of finding soil with good oil, P(FG) = 0.16

The probability of finding medium quality oil, P(FM) = 0.64.

3 0

2 years ago