Combine like terms.<br> 2x^2+7+6x-1-7x-6x^2+1<br> 2x <br> 2<br> +7+6x−1−7x−6x <br> 2<br> +1

George sold 18, 22, 26, 12, 25, 20, and 19 cars per month over the past seven months. He followed the steps below to

ikadub [295]

Answer:

George made the first mistake by multiplying by 7 instead of 8

Step-by-step explanation:

The total cars needed to have a mean of 24:

24 x 8 = 192

Total cars sold during the seven months:

18 + 22 + 26 + 12 + 25 + 20 + 19 = 142

Subtract the total cars sold from the total cars needed:

192 - 142 = 50

George needs to sell 50 cars in the eight month in order to have an average of 24 cars sold per month.

8 0

3 years ago

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The sum of two integers is 30 and their difference is 2

GalinKa [24]

14 and 16 f4gg0t!!!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

Factor the polynomial 54c3d4 + 9c4d2

miv72 [106K]

54c^3d^4 + 9c^4d^2
9c^3d^2(6d^2 + c)

7 0

3 years ago

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What happens to the value of the expression 35 + kas k decreases?

ExtremeBDS [4]

Answer:

a i think

Step-by-step explanation:

3 0

3 years ago

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Combine like terms. 2x^2+7+6x-1-7x-6x^2+1 2x 2 +7+6x−1−7x−6x 2 +1