Answer: Does not exist.
Step-by-step explanation:
Since, given function, f(x) = 6x tan x, where −π/2 < x < π/2.
⇒ f(x) = 
And, for vertical asymptote, cosx= 0
⇒ x = π/2 + nπ where n is any integer.
But, for any n x is does not exist in the interval ( -π/2, π/2)
Therefore, vertical asymptote of f(x) where −π/2 < x < π/2 does not exist.
1 pound is 16 ounces
32 ounces is 32 ÷ 16 = 2 pounds
1 pound of Botcoil ⇒ 1.5
2 pounds = 1.5 × 2 = 3
The answer is 6 units and 3 fourths
You didn't give the fourth zero, but the answer is still false. If you have a root or an imaginary number as a zero, then its conjugate is also a zero. So if 8i is a zero, then -8i must also be a zero, and if 4i is a zero, then -4i must be a zero, with those zeros and -4, the number of zeroes exceeds the number of zeroes that a fourth degree polynomial can have.
1. distribute a 1 to the 12 and 8a
2. you'll get 15+ 12 +8a
3. add the like terms(15 and 12)
4. your final answer will be 27+8a
