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3 years ago

Need to know if I got the correct answer for #29. If it’s wrong can you help me solve it please and thank you.

Mathematics

1 answer:

Pie3 years ago

6 0

Answer:

-x² -2x -5 -12/(x-2)

Step-by-step explanation:

The attachment shows the synthetic division. You can find the remainder by evaluating the dividend for x=2:

-2³ -2 -2 = -12 . . . . matches the above answer, not [A] or [C] shown

_____

You can always multiply out your answer to see if it matches the problem.

(-x^2 -2x +3)(x -2) +3 = -x^3 +7x -3 . . . . not the dividend you started with

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Can someone round this to the nearest tenth

Dominik [7]

Answer:

I think its A

Step-by-step explanation:

I did the math and thats what i got even rounded up

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2 years ago

Is 3/4 less than 1/2 or more than 1/2?

nalin [4]

Answer: more than 1/2

Step-by-step explanation: 1/2 equals 2/4 so 3/4 is more.

4 0

3 years ago

Read 2 more answers

Of the 50 plays attributed to a playwright, 11 are comedies, 19 are tragedies, and 20 are histories. If one play is selected

Cerrena [4.2K]

The answer is 3/5, or 60%

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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

Which of the following best describe(s) the diagonals of a rectangle. Select all that apply.

Evgesh-ka [11]

Answer:

I believe it is congruent and intersecting

Step-by-step explanation:

6 0

3 years ago