This has both a horizontal and a vertical asymptote. There are no slant (oblique) asymptotes cuz the degree of the numerator is not higher than that of the denominator. If the degree of the numerator is less than the degree of the denominator, which is our case here, then the horizontal asymptote is 0. But we also have a vertical asymptote, which occurs where the denominator = 0. We all know that we break every rule known to mankind if we try to divide by 0, so there also a vertical asymptote at x = 2.

7 0

3 years ago

Find the value of x.

Cerrena [4.2K]

$x^2 + (x + 3)^2 = (\sqrt{117})^2$

$x^2 + x^2 + 6x + 9 = 117$

$2x^2 + 6x -108 = 0$

$x^2 + 3x - 54 = 0$

$(x + 9)(x - 6) = 0$

$x + 9 = 0 ~~or~~ x - 6 = 0$

$x = -9 ~~ or~~ x = 6$

Since the side of a triangle cannot have a negative length, we discard the solution x = -9.

Answer: x = 6

3 0

3 years ago

Read 2 more answers

Tom had 8 cookies, and he ate one and a half cookies. how many cookies were left?

Alex777 [14]

Tom had 6 and a half cookies

6 0

3 years ago

Whose responsibility is it to call out the government and point out flaws

Jlenok [28]

Adults, 18 and older

7 0

3 years ago