4m-3n=8, for m wats yeah

3•3^-1 Show work plz!

andreyandreev [35.5K]

Answer:

1

Step-by-step explanation:

Firstly we need to do the exponent,3^-1, which equals 1/3.

Now, we multiply 3 with 1/3.

3•1/3 =1

8 0

3 years ago

Which unit would be the best for measuring the amount of water in a bathtub?

VashaNatasha [74]

The unit that would be the best for measuring the amount of water in a bathtub would be up to the scale used. Out of the given choices, liters seems to be the best choice. So, the correct answer will be C.

8 0

3 years ago

Read 2 more answers

The annual interest rate of Marta's savings account is 7.2% in a simple interest is calculated quarterly. What is the periodic i

Delvig [45]

1. You have the following information:

-The annual interest rate is 7.2%.
- The simple interest is calculated quarterly.

2. Then, to solve this exercise and calculate the periodic interest rate of Marta's Account, you only have to divide the annual interest rate (7.2%) by 1/4 year. So, you have:

=(7.2%)(1/4)
=(7.2%)/4
=1.8%

What is the periodic interest rate of Marta's Account?

The answer is: 1.8%

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B

Margaret [11]

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.

5 0

2 years ago

Cam you look at this picture. I really need help for #2, #5, #8, #9, #10

jasenka [17]

# 2 is true
# 5 is 0.375
#8 is 8 solutions
#9 is %25
#10 is %37.5

hope this helps

for number 2: 4 times 4 equals 16 is that's why its the square root of 16.
for number 8: with red there are 3 options plus the red option so that's 4 plus the same thing with blue so that's 8.
for number 9: there is a 2/8 chance of getting that. 2/8= 25%
for number 10: there is a 3/8 chance of landing on it. 3/8= 37.5%

3 0

3 years ago