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algol13
3 years ago
8

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?
Mathematics
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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Eva8 [605]

Answer:

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Step-by-step explanation:

7 0
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)

6 0
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
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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
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