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.
I think the answer might be c
7) 8.9
8) 6.7
9) 12.2
10) 7.6
11) 7.3
12) 6.3
Elimination:
-4x+y=6
-5x-y=21 ...and then u add them together so its;
(the y cancelled out so its 0) x =27
plug in the 27 to any x...
-4(27) + y = 6
-108+y=6
add -108 on both sides so its...
y= -102
(27, -102)
Hope this helped :)
→ a
the equation is y = mx ( where m is the slope / constant of variation )
calculate m using the gradient formula
m = ( y₂ - y₁ ) / (x₂ - x₁ )
with (x₁, y₁) = (0, 0 ) and (x₂, y₂ ) = (4, 1) ← 2 points on the line
m =
= 