Answer:
Remove all perfect squares from inside the square root. ... I think it's about eighth or ninth grade. ... so if you have the cube root of the square root of (x-5) =2, you get ((x-5)^(1/2))^1/3 = 2, power to power requires multiplication, so (x-5)^1/6 = 2, ...
Missing: 176 xy
Store one has 15 rolls for 3.60 so divide it answer will be .24 per roll. store 2 has 18 rolls for 3.80. divide it up answer is .21 per roll. so store 2 is the answer.
1) function f(x)
x - 5
f(x) = ----------------
3x^2 - 17x - 28
2) factor the denominator:
3x^2 - 17x - 28 = (3x + 4)(x - 7)
x - 5
=> f(x) = -----------------------
(3x + 4) (x - 7)
3) Find the limits when x → - 4/3 and when x → 7
Lim of f(x) when x → - 4/3 = +/- ∞
=> vertical assymptote x = - 4/3
Lim of f(x) when x → 7 = +/- ∞
=> vertical assymptote x = 7
Answer: there are assympotes at x = 7 and x = - 4/3
3-4-2-1-8-1-2-9 answer is not ok im not sure
Answer:
You would be correct!
Step-by-step explanation:
10 × 4 = 40 40 + 15 = 55 therefore D is the answer great job!