Skip count by twos.<br> 30,_ _ _ _ _ _ _ _

alex41 [277]

Answer:

125

Step-by-step explanation:

Steps to Solving the Problem

So let's say you have five friends who who are each going to pay you five dollars for the next five days. You can figure out how much money you will receive by calculating five cubed or five to the third power. This will tell us how much money you will get because you will multiply 5 by 5 by 5 or 5 cubed.

To start, you need to turn words into numbers by figuring out the base and exponent. An exponent is the number after the ^ which tells you how many times to multiply the base number by itself. The base number comes before the ^. When five is cubed, five is the base number which will show up before the ^.

Three is the exponent and will be the number after the ^. We know three is the exponent because of the word 'cubed,' which always represents the number three.

We can write out five cubed as:

5^3

Once the exponent and base are figured out, we need to write out the long version, which shows all the multiplication being done. In this case, the exponent of three tells us to multiply five by itself three times.

5 x 5 x 5

This shows us 5 friends multiplied by $5 multiplied by 5 days.

From here, multiply the numbers out to get the final answer. Multiply two numbers at a time to so you don't make a mistake.

5 x 5 = 25

25 x 5 = 125

6 0

3 years ago

A customer pays $3.27 for oranges and $4.76 for pears. How many pounds of fruit does the customer buy?

ruslelena [56]

Take 3.27 and divide that by 1.09 (for the oranges)
3.27/1.09=3lbs. of oranges
then take 4.76 and divide that by 1.19 (for the pears)
4.76/1.19=4lbs. of pears
add 4+3 to get how many lbs. in all
4+3=7

3 0

3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$