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2 years ago

Please help need answer fastwill give brainliest

Mathematics

1 answer:

ozzi2 years ago

4 0

Ummm i dont know but i think its c

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A volleyball of radius 13 inches fits exactly in a cylinder with the same radius. inside the container there is empty space not

asambeis [7]

Volume of Sphere is V=(4/3)*PI*r^3

The radius of the volleyball is 13 inches

<span>So V=(4/3) * 3.14 * 13^3 = 9198 cubic inches</span>

Volume of a cylinder is V=pi * r^2 *h

Since they say the ball fits perfectly

The radius is 13 and the height is 26 ( the diameter of the ball)

So V=3.14 * 13^2 * 26 = 13,797 cubic inches

13797 – 9198 = 4,599 cubic inches of empty space

4 0

3 years ago

Read 2 more answers

Cain measured his couch and found it is 3 3/4 meters long what is the length as a decimal

11Alexandr11 [23.1K]

3 3/4
= 3+ 3/4
= 3+ 0.75
= 3.75

The length as a decimal is 3.75 m~

5 0

3 years ago

At a game booth, a student gets a box of candy as the prize for winning a game. The boxes come in four colors: white, red, green

PSYCHO15rus [73]

Your answer is 6 over 24 multiplied by 6 over 23.

4 0

2 years ago

All right angles are congruent. Converse Inverse: Contrapositive

WITCHER [35]

Answer:

No not all right angles are congruent.

Step-by-step explanation:

7 0

3 years ago

Find the volume of the following solid. The solid between the cylinder f(x,y)equals=e Superscript negative xe−x and the region

PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy$

Now evaluate,

$\int_{0}^{ln(6)}e^{-x}dx$

which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

$\int_{-6}^{6}\frac{5}{6}dy = 10$

So,

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10$

7 0

2 years ago