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3 years ago

Thank you for taking the time to look at my question

Mathematics

2 answers:

Deffense [45]3 years ago

8 0

Answer:

I think it's B but im not sure

hope this helps

have a good day :)

Step-by-step explanation:

k0ka [10]3 years ago

5 0

Answer and Step-by-step explanation:

The functions f(x) is being put into the function of g(x).

$\frac{2x + 3}{4}$

$\frac{1}{2} x + \frac{3}{4}$

Our answer is Answer choice 2.

#teamtrees #PAW (Plant And Water)

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Solve for x in the equation x 2 - 4 x - 9 = 29.

Marizza181 [45]

Answer:

x= -19

Step-by-step explanation:

2x-4x-9=29

-2x=29+9

x=38/-2

= -19

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3 years ago

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Zoey makes $75 for 3 hours of work. How much money did Zoey make each hour?

stiks02 [169]

75/25 = 3
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2 years ago

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What is the first step when evaluating the expression 8 + 6 – 4 ÷ 4 + 2?

Juliette [100K]

Divide four by four first

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3 years ago

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Algebra 2 need help on a ( ABC )

Len [333]

Answer:

$x = \frac{15}{4}$

Step-by-step explanation:

b) set up equations so they are equal to each other,

$2x - 2 = - 3x + 12$

$5x = 14$

$x = \frac{14}{5}$

this is when f(x)=g(x) so our approximation was close.

c)solving it graphically is nearly impossible because the solution can be any value around the intersection. only way to be sure is to solve it symbolically.

3 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago