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3 years ago

True or False: The following pair of functions are inverse functions.

Mathematics

2 answers:

AURORKA [14]3 years ago

8 0

Answer:

False is the correct answer✔✔✔

Step-by-step explanation:

Ira Lisetskai [31]3 years ago

3 0

Answer:

its is false

but yeah just adding tk it

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A coat that usually costs $120.00 is marked 1/3 off. What is the sale price of the coat?

dedylja [7]

If the coat is 120$ then 1/3 is taken off it would be 80$

Step-by-step explanation:

1/3 of 120 is 40. 40x 3 is 120.

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3 years ago

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A bicycle has two wheels.how many terms do u think a binomial has

Alika [10]

A binomial is going to have two terms. =)

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3 years ago

Order the fractions from smallest to largest. 1\4 1\8 1\2 1\5

sukhopar [10]

The answer is pretty simple, 1/8, 1/5, 1/4, 1/2

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3 years ago

Misty, Tina and Jen walk into a store. The manager places his hand on Tina’s shoulder and tells her, “You didn’t win the door pr

vladimir1956 [14]

Answer:

50%

Step-by-step explanation:

3 of them walk into a store, giving her a 33.333% chance of winning. By eliminating one, Tina may as well have never walked into the store, leaving Jen with a 50% chance.

3 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago