All categories

4 years ago

36x^3a+45xa^2=9xa( ) complete the statement.

1 answer:

6 0

9xa is the common factor.

You can complete the statement by dividing the two terms of the expression of the left side by 9xa, and you will get:

36x^3a + 45xa^2 = 9xa ( 4x^2 + 5a)

You might be interested in

An angle co-terminal with 5pi/7 is: a) -9pi/7 b) -2pi/7 c) 12pi/7 d) 20pi/7

just olya [345]

To find co-terminal angles, add or subtract 2π. Make sure the denominator is the same as the original problem.

5π/7
+ 14π/7
19π/7

5π/7
- 14π/7
-9π/7

a

5 0

3 years ago

Why do you look so weird

muminat

Answer:

cause i'm built different:)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help please help

kykrilka [37]

Answer:

3 rounds

Step-by-step explanation:

Variable x = number of rounds

2x + 13 = 3x + 10

13 = x + 10

3 = x

5 0

3 years ago

You have $2 and earn $0.50 for each cup of lemonade you sell. Write an equation in two variables that represents the total amoun

Colt1911 [192]

A= 2+0.50c this is your equation. A being your dependent variable and c your independent variable. Hope this helps. :)

7 0

3 years ago

Oliver invested $970 in an account paying an interest rate of 7 1/2 % compounded continuously. Carson invested $970 in an accoun

hjlf

Answer:

0.50 or about half a year longer.

Step-by-step explanation:

We can write an equation to model bot investments.

Oliver invested $970 in an account paying an interest rate of 7.5% compounded continuously.

Recall that continuous compound is given by the equation:

$A = Pe^{rt}$

Where A is the amount afterwards, P is the principal amount, r is the rate, and t is the time in years.

Since the initial investment is $970 at a rate of 7.5%:

$A = 970e^{0.075t}$

Carson invested $970 in an account paying an interest rate of 7.375% compounded annually.

Recall that compound interest is given by the equation:

$\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}$

Where A is the amount afterwards, P is the principal amount, r is the rate, n is the number of times compounded per year, and t is the time in years.

Since the initial investment is $970 at a rate of 7.375% compounded annually:

$\displaystyle A = 970\left(1+\frac{0.07375}{1}\right)^{(1)t}=970(1.07375)^t$

When Oliver's money doubles, he will have $1,940 afterwards. Hence:

$1940= 970e^{0.075t}$

Solve for t:

$\displaystyle 2 = e^{0.075t}$

Take the natural log of both sides:

$\ln\left (2\right) = \ln\left(e^{0.075t}\right)$

Simplify:

$\ln(2) = 0.075t\Rightarrow \displaystyle t = \frac{\ln(2)}{0.075}\text{ years}$

When Carson's money doubles, he will have $1,940 afterwards. Hence:

$\displaystyle 1940=970(1.07375)^t$

Solve for t:

$2=(1.07375)^t$

Take the natural log of both sides:

$\ln(2)=\ln\left((1.07375)^t\right)$

Simplify:

$\ln(2)=t\ln\left((1.07375)\right)$

Hence:

$\displaystyle t = \frac{\ln(2)}{\ln(1.07375)}$

Then it will take Carson's money:

$\displaystyle \Delta t = \frac{\ln(2)}{\ln(1.07375)}-\frac{\ln(2)}{0.075}=0.4991\approx 0.50$

About 0.50 or half a year longer to double than Oliver's money.

6 0

3 years ago