The domain { x | x = -5 , -3 , 1 , 2 , 6}
17) f(x) = 16/(13-x).
In order to find domain, we need to set denominator expression equal to 0 and solve for x.
And that would be excluded value of domain.
13-x =0
Adding x on both sides, we get
13-x +x = x.
13=x.
Therefore, domain is All real numbers except 13.
18).f(x) = (x-4)(x+9)/(x^2-1).
In order to find the vertical asymptote, set denominator equal to 0 and solve for x.
x^2 -1 = 0
x^2 -1^2 = 0.
Factoring out
(x-1)(x+1) =0.
x-1=0 and x+1 =0.
x=1 and x=-1.
Therefore, Vertical asymptote would be
x=1 and x=-1
19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)
We have degrees of numberator and denominator are same.
Therefore, Horizontal asymptote is the fraction of leading coefficents.
That is 7/2.
20) f(x)=(x^2+3x-2)/(x-2).
The degree of numerator is 2 and degree of denominator is 1.
2>1.
Degree of numerator > degree of denominator .
Therefore, there would no any Horizontal asymptote.
Question: solve for r: 6 ^ (2r) = 800
Answer: 1.87
Step-by-step explanation:
6 ^ (2r) = 800
log(6 ^ (2r)) = log(800)
2r log(6) = log(800)
r = log(800) / (2 log(6))
r = 1.86537641979
round
r = 1.87
log can be any logarithm function.
6^(2×1.86) = 784.734328546
6^(2×1.87) = 813.365367336
Answer: 16
Step-by-step explanation:
multiplying by negative exponents can be a little tricky. So (1/4x)^-2 becomes 16/x^2 = __/x^2 so the missing number is 16
