Ask question

Ask question

All categories

2 years ago

14

Eden opened a new checking account on Monday and deposited $25 into it. The table below shows the activity in her account for th

e week. Use this information to determine which of the following statements are true Select two that apply.

1 answer:

exis [7]2 years ago

5 0

If I should solve i need the table given to you

You might be interested in

Khan khan khan khan

Ber [7]

Answer:

11

Step-by-step explanation:

it says motor cycles combined with garbage so add both of them then subtract that from the bull dosers

motor =18

garbage=4

18+4=22

bulldosers=11

22-11= 11 more

5 0

2 years ago

Read 2 more answers

please help me with it and dont just put random letters cause i dont know why u do it its not like u can buy a house with these

BartSMP [9]

Answer:

v=432cm

sA=144cm

Step-by-step explanation:

1)))

v=w*d*h

v=3*9*16

v=432cm

2)))

sA=h*w

sA=16*9

sA=144cm

6 0

2 years ago

Read 2 more answers

A company can manufacture x hundred items for a total cost of C=300+1200x-100x. Find x if the total cost is 3,000

Scilla [17]

C=300+1200x-100x. Find x if the total cost is 3,000
3,000 = 300 + 1200x -100x
3000 = 300 + 1100x
2700 = 1100x
2700÷1100=x
x=27/11
x = 2 5/11 Round to 3

5 0

2 years ago

Find the sale price of each item. Round to two decimal places when necessary.

Bond [772]

Answer:

Sale Price: 276.25

Step-by-step explanation:

The original price of the item is $325 and the markdown is 15% off.

So, all you have to do is find out what 15% of $325, which it's 48.75, and subtract that from $325 to get your final sale price, which is $276.25.

7 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

5.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

8.

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago