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3 years ago

13

Each edge of a cube measures 12.1 in. What is the surface area of the cube?

1 answer:

REY [17]3 years ago

8 0

878.46 in. is your answer! Hope this helps!

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The ladder in a fire station is 29 feet (ft) tall. The ladder is anchored to the basement floor and the second floor ceiling. Th

Eduardwww [97]

Answer:

top = 17ft

bottom = -12ft

Step-by-step explanation:

As seen in the badly drawn picture attached in this question we are trying to find the point at the top of the ladder and the point at the bottom of the ladder. As seen in the picture ground level has an altitude of 0 ft meaning that lower than this would be in the negatives. Since the basement floor is 12 feet below ground level then this point would be -12 ft. Now since we know that the ladder is 29ft tall we simply add this height to the basement floor value to get the value of the 2nd floor ceiling (top of the ladder).

29ft + ( -12ft) = 17ft

Therefore we can see that the point at the top of the ladder is 17ft.

8 0

3 years ago

It takes Jada 20 minutes to walk to school. It takes Andre 80% as long to walk to school. How long does it take Andre to walk to

Scrat [10]

Answer:

16 minutes

Step-by-step explanation:

If it takes Jada 20 minutes and it takes Andre 80% as long, then you multiply 20 by %80

20 * .80 = 16 minutes

7 0

2 years ago

Read 2 more answers

If there are 93 calories in 3/4 cups how many calories are in 1 cup

liq [111]

93 = 3/4
?? = 1
Cross multiple
1(93)/(3/4) = 124

4 0

2 years ago

Read 2 more answers

2 dogs, 4 horses 1 giraffe and a duck are lying on the bed. 3 chickens are flying over a chair. How many legs are on the ground?

galben [10]

Answer:

10

Step-by-step explanation:

2 legs because you walked into the room and 4 legs of the bed and 4 legs of the chair.

4 0

1 year ago

Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

5 0

3 years ago