1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kvv77 [185]
2 years ago
14

What is the volume? 2 cm 2 cm 2 cm cubic centimeters

Mathematics
2 answers:
Aliun [14]2 years ago
3 0

Answer:

V = 8 cm³

Step-by-step explanation:

the volume (V) of a cube is calculated as

V = s³ ( s is the side length )

here s = 2 , then

V = 2³ = 8 cm³

lawyer [7]2 years ago
3 0

Answer:

8 cm

Step-by-step explanation:

<h2>length = 2 cm</h2><h2>breadth = 2 cm</h2><h2>height = 2 cm</h2><h2>volume = L*b*h</h2><h2>              = 2*2*2</h2><h2>               = 8 cm</h2>

You might be interested in
Which line segment shows the base that corresponds to the given height of the triangle?
maksim [4K]

Answer:

A

Step-by-step explanation:

The height of the triangle is always perpendicular to the base. Therefore, in this figure, the base of the triangle corresponds to segment A

8 0
3 years ago
Read 2 more answers
Which of the following would be equivalent to 92 ⋅ 96?
Elina [12.6K]
The answer is
Solution: 9^10 over 9^2
3 0
3 years ago
Read 2 more answers
Xy''+2y'-xy by frobenius method
aalyn [17]
First note that x=0 is a regular singular point; in particular x=0 is a pole of order 1 for \dfrac2x.

We seek a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^{n+r}

where r is to be determined. Differentiating, we have

y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}
y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}

and substituting into the ODE gives

\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0

The indicial polynomial, r(r+1), has roots at r=0 and r=-1. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When r=0, we have the recurrence

a_n=\dfrac{a_{n-2}}{(n+1)(n)}

valid for n\ge2. When n=2k, with k\in\{0,1,2,3,\ldots\}, we find

a_0=a_0
a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}
a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}
a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}

and so on, with a general pattern of

a_{n=2k}=\dfrac{a_0}{(2k+1)!}

Similarly, when n=2k+1 for k\in\{0,1,2,3,\ldots\}, we find

a_1=a_1
a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}
a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}
a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}

and so on, with the general pattern

a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}

So the first indicial root admits the solution

y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}

which you can recognize as the power series for \dfrac{\sinh x}x and \dfrac{\cosh x}x.

To be more precise, the second series actually converges to \dfrac{\cosh x-1}x, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When r=-1, we may seek a second solution of the form

y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n

where y_1=\dfrac{\sinh x+\cosh x-1}x. Substituting this into the ODE, you'll find that c=0, and so we're left with

y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n
y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots

Expanding y_1, you'll see that all the terms x^n with n\ge0 in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form y_2=\dfrac1x. Adding this to y_1, we end up with just \dfrac{\sinh x+\cosh x}x.

This means the general solution for the ODE is

y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x
3 0
3 years ago
[X + y] + 3w? Helpppppp
lana [24]

Answer:

x+y+3w

Step-by-step explanation:

7 0
3 years ago
The Ratio of two numbers is 3/5, and their sum is 104. What are the 2 numbers?
Rom4ik [11]

Answer:

39 and 65

Step-by-step explanation:

the ratio of the two numbers is 3 : 5

sum the parts of the ratio, 3 + 5 = 8 parts

divide the sum by 8 to find the value of one part of the ratio

104 ÷ 8 = 13 ← value of 1 part of the ratio , then

3 parts = 3 × 13 = 39

5 parts = 5 × 13 = 65

the two numbers are 39 and 65

4 0
2 years ago
Other questions:
  • marco is 6 feet 3 inches tall. There are 2.54 centimeters in 1 inch. What is marcos height in centimeters.
    10·2 answers
  • I'd really appreciate it if anyone could help! :) I'll give Brainliest!
    5·1 answer
  • Use f(x) = 1/2x and f -1(x) = 2x to solve the problems.
    7·1 answer
  • WHO CAN HELP AT LEAST ONEBOF THEM I NEED HELP ASAP
    8·1 answer
  • A race organizer brought 24 gallons of sports drink for a race. How many quarts of sports drink did he buy?
    15·2 answers
  • What is the correct unit of measurement to use when filling up a tub of water?
    10·1 answer
  • Help Me Please! And Hurry!<br>​
    5·2 answers
  • Solve the inequality: 9x-4x+4≥36-12
    14·2 answers
  • Please help me pleaseeee
    12·1 answer
  • On a recent flight, the 120 passengers ordered dinner as depicted in the table. Use this table to answer the following question.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!