G(n) = 2n + 3 h(n)=n-1 Find (g•h)(n)

Use synthetic division to find the result when 3x^3 + 13x^2 + x – 12 is divided by X + 4.

IgorC [24]

Answer:

3x² + x - 3

Step-by-step explanation:

The way I like to do it is to get rid of the x's and just use the numbers in the case of synthetic division

Pretend the square root of the division symbol

-4 √ 3 12 1 -12

Quick note: We're dividing by -4 because we're dividing by x + 4

First bring down the 3

-4 √ 3 13 1 -12

3

Multiply it by 4 then bring add it to the next number

-4 √ 3 13 1 -12

-12

3 1

Add that number to the next one

-4 √ 3 13 1 -12

-12 -4

3 1 -3

Finally repeat the step for the last number

-4 √ 3 13 1 -12

3 -12 -4 +12

3 1 -3 0

Now take those bottom numbers and add back the x's but with one less power, so the starting x³ would now become x², x² would become x, and so on

3x² + 1x - 3

8 0

2 years ago

A box contains 8 green balls, 16 red balls, 10 white balls, and 6 pink balls.

Pepsi [2]

The probability of drawing a white ball is 1/4

Explanation:
When we solve for probability, we want to find out how many of something there is out of its total. So in this case, we want to see how many white balls there are out of the totals balls in the box. We already know that there are 10 white balls, so we need to figure out the total balls in the box, by adding up. 8 green balls + 16 red balls + 10 white balls + 6 pink balls = 40 balls in total. Now we know that the fraction would be 10/40, as there are 10 white balls out of 40 total balls. Finally, we simplify the fraction to 1/4, therefore the answer is B.

8 0

3 years ago

Read 2 more answers

Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.

zloy xaker [14]

The equation for cosine is cos(x)=Adjacent/Hypotenuse
The inside trig function is arccos(3/5), which means cos(x)=3/5. Comparing cos(x)=Adjacent/Hypotenuse with cos(x)=3/5

Find Adjacent=3 and Hypotenuse=5.
Then, using the Pythagorean theorem, find Opposite=?
a² = c² - b²
a² = 5² - 3² = 25 - 9 = 16
a = √16 = 4

Adjacent=3Opposite=4Hypotenuse=5

Plug in the value for sin(x) = opposite/hypotenuse

sin(x) = 4/5

6 0

3 years ago

3 lines are shown. A line with points C, A, F intersects a line with points E, A, B at point A. A line extends from point A to p

Verizon [17]

Answer:

m < EAD = 29 degrees

m < CAB = 119 degrees

Given :

The question states that m < CAE = m<FAB = 61 degrees and m<DAF = 90 degrees

Solution:

1. Since line CAF and EAB intersect each other, m<CAF = m< EAF - (opposite vertical angles are equivalent)

2. m<BAC + m<EAC = 180 degrees (sum of linear pair)

3. m<CAB = 180 degrees - m<EAC

4. Equation 1: m<CAB = m<EAF = 119 degrees

5. m<EAF = m<EAD + m< DAF

6. m<EAD = m<EAF - m<DAF

7. m<EAD = 119 degrees-90 degrees = 29 degrees

Hope this helps!!! :)

Attached photo work is created by another brainly user named tallinn.

Click on this link to find his work and possibly more information of the answer. Thanks!

5 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