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3 years ago

Please Answer this question 15 points

Mathematics

2 answers:

WINSTONCH [101]3 years ago

5 0

Answer:

btw it olny 8 points but it is c

Step-by-step explanation:

Mashcka [7]3 years ago

4 0

Answer:

Step-by-step explanation:

to find (f/g)(x), you have to divide f(x) by g(x)

$\frac{\frac{(x+16)}{x}}{\frac{(2x)}{(x+4)} } = \frac{(x+16)}{x} * \frac{(x+4)}{2x} = \frac{(x+4)(x+16)}{2x^2}$

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Given: f(x) = 3x-7, g(x)=2x2 -3x+1, h(x)=4x+1, k(x) =-x2+3 Find: k(x) +g(x)

Afina-wow [57]

Answer:

k(x) + g(x) = x² - 3x + 4

General Formulas and Concepts:

Combining like terms

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x - 7

g(x) = 2x² - 3x + 1

h(x) = 4x + 1

k(x) = -x² + 3

Substitute: k(x) + g(x) = -x² + 3 + 2x² - 3x + 1
Combine like terms (x²): k(x) + g(x) = x² + 3 - 3x + 1
Combine like terms (Z): k(x) + g(x) = x² - 3x + 4

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2 years ago

What is the greatest common factor of 18, 36, and 90?

hoa [83]

the answer is 18 yes you are correct

8 0

3 years ago

¿Cómo simplifico 57/10?

Nonamiya [84]

Answer:

Step-by-step explanation:

57/10 cannot be simplified

but the decimal answer is 5.7

3 0

3 years ago

How can you use rates to determine wheather a situation is a proportional relationship

svp [43]

<em>If the rate is a constant, the relationship is proportional.</em> (The rate is the constant of porportionality.)

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For two or more points on the function curve, divide the dependent variable value by the independent one. If you get the same result in every case, the relationship is porportional.

___

For some non-proportional relationships, it is possible to find points on the graph that will pass the above test. If you suspect the relationship is actually not one of proportionality, try more points. Check also to make sure that (0, 0) is on the curve.

7 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