Answer:
true
Step-by-step explanation:
for a right triangle no
Simplifying h(x) gives
h(x) = (x² - 3x - 4) / (x + 2)
h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)
h(x) = ((x + 2)² - 7x - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)
h(x) = (x + 2) - 7 - 22/(x + 2)
h(x) = x - 5 - 22/(x + 2)
An oblique asymptote of h(x) is a linear function p(x) = ax + b such that

In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.
Answer: Choice A
Set the radicand (stuff under the radical) greater than or equal to 0. Solve x+11 >= 0 for x to get x >= -11
Answer:
2 and 5 hundredths
Step-by-step explanation:
3 - 1 = 2
14 hundredths - 9 hundredths = 5 hundredths
hence you get 2 and 5 hundredths
Triangle, rectangles, pollygon, robtuse, i hope this is what you mean