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TiliK225 [7]
2 years ago
13

Find the following for the function f(x) = x/x^2+1

Mathematics
1 answer:
7nadin3 [17]2 years ago
8 0

Step-by-step explanation:

step 1. f(0) = 0/(0^2 + 1) = 0/1 = 0

step 2. -f(x) = -x/(x^2 + 1)

step 3. f(9) = 9/(9^2 + 1) = 9/82.

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What is an equation of a line, in point-slope form, that passes through (1, −7) and has a slope of −2/3
Firdavs [7]

Answer:

D+B 12

Step-by-step explanation:

4 0
3 years ago
I will give Brainliest
ioda

Answer:

The volume(V) of gas will increase when the pressure of the gas decreases assuming all the other variables are held constant.

Step-by-step explanation:

 P = Pressure of the ideal gas

            V = Volume of the ideal gas

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4 0
3 years ago
Read 2 more answers
Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

8 0
3 years ago
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5 0
3 years ago
Read 2 more answers
Mt. McKinley, in Alaska, is the highest mountain in North America at 20,320
gladu [14]

Answer:

12700 feet

Step-by-step explanation:

Do 5/8 of 20,320

Do 20320 divided by 8 which is 2540

Do 2540 times 5 which is 12700

6 0
2 years ago
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