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2 years ago

Can someone please answer this ASAP??

Mathematics

1 answer:

loris [4]2 years ago

5 0

V=πr²h/3
Diameter is 24ft, Radius = 12 ft
V=π12²10/3≈1507.96

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What is the plants height after 20 minutes?

ICE Princess25 [194]

Answer:

14.60

Step-by-step explanation:

Note that there are 100 centimeters in 1 meter. The plant is 13 meters tall, and grows 8 centimeter every month. It is asking you for 20 months worth (x = 20) of growth.

First, solve for the amount of centimeters total in 20 months:

8 centimeter per month x 20 months = 160 centimeters per 20 months.

Remember, there are 100 centimeters in every meter. Therefore:

160 centimeter/ 100 centimeters per meter = 1.60 meters.

Add the growth to the original height:

13.00 + 1.60 = 14.60

14.60 meters is your answer.

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2 years ago

All equilateral triangles are isosceles? A. True B. False

Maru [420]

Answer:

A. true

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which of the following is a rational number? a)10 b)100 c)1000 d)100000

earnstyle [38]

Answer:

All of the above

Step-by-step explanation:

10, 100, 1000, 100000 all are rational numbers because you can write them as:

10/1 , 100/1, 1000/1, 100000/1

7 0

2 years ago

It’s on the picture .

Marrrta [24]

Answer:

slope 2

yint 1

equation y=2x+1

3 0

3 years ago

Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov

mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

$2t=(m+\frac{1}{2} ).\frac{\lambda}{n}$

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

$t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm$

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

$t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm$

Hence the thickness difference should be

$t_1-t_2=131.58-75.19=56.39 \ nm$

7 0

2 years ago