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GREYUIT [131]
3 years ago
10

The surface areas of two similar figures are 25in^2 and 36in^2. If the volume of the smaller figure is 250in^3, what is the volu

me of the larger figure?
Mathematics
1 answer:
Strike441 [17]3 years ago
3 0
What you would do is add 25in^2 and 36in^2 together to get your answer. thanks for asking a question I could answer.
You might be interested in
A driveway 18 ft wide and 36 ft. long is to be paved with concrete 3 in. thick. How many cubic yards of concrete are required?
amm1812

Answer:

6 yds^3

Step-by-step explanation:

I find these easiest if you put all of the dimensions into yards and then multiply

18 ft =  6 yds

36 ft = 12 yds

3 in = 3/36 yds    = 1/12 yds     now multiply them together

6 x 12 x 1/12 = 6 yds^3

8 0
2 years ago
C. If one cubic foot contains about 7.5 gallons, how many gallons of water are in the pool? (3 points) I Water: gallons 3. A rec
Dominik [7]

Answer:

<u>One Rectangular Pan</u>

Step-by-step explanation:

Part C cannot be answered since the dimensions of the pool are not given.  Please check the question.

The cake pans may be compared by calculating the volume of each shape:

Rectangular pan:  Volume = Length x Width x Height

  Volume Rectangle = (13")*(9")*(2") = <u>234 in^3</u>

Round pan:  Volume = \pi r^{2} h

  Volume Round = (3.14)*(4")^2(2") = 100.5 in^3

Two round pans will hold 2*(100.5 in^3) or 201 in^3 of batter

One rectangular pan holds 234 in^3.

Go with the rectangular pan if you want more cake.

4 0
1 year ago
Mr. Gill is draining his pool. The pump he is using changes the water level by −2 1/4 inches per hour. A stronger pump would dra
professor190 [17]

Answer:

-5.625 in/hour

Step-by-step explanation:

−2 1/4 * 2 1/2 = -5.625 in/hour

7 0
3 years ago
A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

velocity = frequency × wavelength   .....................2

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × \rho × g = mg

and

T = v × \rho - v × \rho × g    

from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
3 years ago
NEED HELP PLEASE!!! Please explain your answer!
Zanzabum
             <span> You must make it a fraction by multiplying by 100, moving the decimal over by 2 spaces right so we have 36. Then, put it over 100, because it is a fraction. Now, with 36/100 we must simplify by dividing each number by another # that goes into it evenly, such as 2. We get 18/50, reduce by 2, we get 9/25. There you go hope I helped </span>
7 0
2 years ago
Read 2 more answers
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