QUICK NOW!!!!!!! Brainliest awarded to the best.

Mathematics

1 answer:

vazorg [7]3 years ago

8 0

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

You might be interested in

If 0 < x < 1 and 0 < y < 1, which of the following must be true?

katovenus [111]

0<x<1 and 0<y<1

x>0 so x is positive and y>0 so y is also positive.
When you multiply two positive numbers you always get a positive number, so the product of x and y must be positive, or greater than 0.
xy>0 - it must be true
xy<0 - it can't be true
Also when you divide a positive number by a positive number you always get a positive number, so the quotient of x and y must be positive.
x/y<0 - it can't be true
D and E can be true, but don't have to. It depends on the values of x and y. If x>y, then x-y>0 is true and x-y<0 isn't true; if x<y, then x-y>0 isn't true and x-y<0 is true.

Therefore, only A must be true.

3 0

3 years ago

Which describes the behavior of the function f(x)= tan^-1(x)?

Fantom [35]

Answer:

Third Option

Step-by-step explanation:

We know that the function $tan(x)$ is defined as $y=\frac{sin(x)}{cos(x)}$ . Since the denominator is $cos(x)$ then we know that $cos(x)=0$ when $x=\frac{\pi}{2}$

We also know that the division by 0 is not defined. Therefore, the limit of $y=tan(x)$ when "x" tends to $\frac{\pi}{2}$ is infinite.

The function $tan^{-1}(x)$ is the inverse of $tan(x)$

By definition, if we have a function f(x), its domain will be equal to the range of its inverse function $f^{-1}(x)$ . If $f(3)=8$ , then $f^{-1}(8)=3$