Where are the asymptotes for the following function located? f (x) = StartFraction 14 Over (x minus 5) (x + 1) EndFraction
1 answer:
Answer:
2 vertical asymptotes occurring at x = 5 and x = -1
Step-by-step explanation:
given
recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)
Equating the denominator to zero,
(x-5)(x+1) = 0
(x-5) =0
x = 5 (first asymptote)
or (x+1) = 0
x = -1 (2nd asymptote)
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Answer:
n¹³
Step-by-step explanation:
Division and multiplication have the same priority in pemdas, so you can solve this by dividing/multiplying from left to right.
Also remember that when dividing exponents of the same base, you subtract them, and when multiplying, you add them.
n⁶· n⁵ · n⁴ ÷ n³ · n² ÷ n
= (n⁶· n⁵) · n⁴ ÷ n³ · n² ÷ n
= (n¹¹ · n⁴) ÷ n³ · n² ÷ n
= (n¹⁵ ÷ n³) · n² ÷ n
= (n¹² · n²) ÷ n
= n¹⁴ ÷ n
= n¹³
Ultimately you just have to calculate
6 + 5 + 4 - 3 + 2 - 1 = 13
Step-by-step explanation:
Using the point C (-4,-2) and the refection rule: