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3 years ago

A cube has a volume of 216cm/3.what is the area of the base of the cube

Mathematics

2 answers:

nika2105 [10]3 years ago

8 0

Agreed with guy above good luck

Anettt [7]3 years ago

7 0

Answer: 36 cm^2

Step-by-step explanation:

Hello!

We can start by finding the cube root of 216, which is 6. This is the length of each side of the cube.

Now, the area of the base of the cube is equal to 6^2=36 cm^2.

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liberstina [14]

Arc length is the angle/360 times circumference. The diameter of the unit circle is 2. So the circumference of the unit circle is 2pi, if you use 3.14 for pi, then the circumference is 6.28. So your equation is
x/360 times 6.28=4.2, divide by 6.28, then multiply by 360
x=240.76432
Your answer rounded to the nearest thousanth is 240.764

6 0

3 years ago

A snail crawled of a centimeter in 2 minutes. At that rate, how far could the snail crawl in 1

vekshin1

Answer:30 centimeters in 1 minute

8 0

3 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

HELP ME FOR BRAIN PLEASE MS SMARTIEZ OR SOMEONE IM CRYING TODAYS MY BIRTHDAY AND ALL I HAVE IS SCHOOL WORK ITS SO TERRIBLE !!

Ket [755]

The answer to the first part of your question is 9
as for the second part 36/4 (which is 9 btw) then subtract 2 I answered this one the best I could (let me know if I'm wrong)

(i just don't have to much time on my hands because I have homework myself)

3 0

4 years ago

..........................answer

Akimi4 [234]

Well it’s simply the answer

8 0

3 years ago

Read 2 more answers