The surface area of one cube is twice the surface area of a second cube.

Mathematics

1 answer:

Usimov [2.4K]3 years ago

4 0

Answer:

$2\sqrt{2} :1$

Step-by-step explanation:

This question was linked from another one regarding the lengths of the cubes, you can find it here: brainly.com/question/22396279

That question gave you the ratio of the lengths, which is $\sqrt{2} :1$ . Now in geometry, a general rule about solids is that if they are similar, the ratio of their volumes is the cube of the ratio of their edges. In a cube's case, every single cube known to mankind is similar as all the edges are the same length.

So the ratio of volumes would be $\sqrt{2} ^3:1^3$ , which can be simplified as $2\sqrt{2} :1$ .

You might be interested in

What are two numbers that when multiplied equal -2, and also when added equals 1

lara [203]

Two numbers that multiply to -2 could be

-1 and 2

Or

-2 and 1.

Since -1+2=1, -1 and 2 are the numbers you are looking for.

Hope this helps!

4 0

4 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$