Answer
Find out the altitude of the equilateral triangle .
To proof
By using the trignometric identity.

As shown in the diagram
and putting the values of the angles , base and perpendicular


solving


As

put in the above
a = 4 × 3
a = 12 units
The length of the altitude of the equilateral triangle is 12 units .
Option (F) is correct .
Hence proved
Step-by-step explanation:
a=2
b=-7
c=-15
-(-7)+-sq root (-7)²-4(2)(-15) / 2(2)
7 +- sq root 169 / 4
7+- 13 / 4
7+13 / 4, 7-13 / 4
x=5 x=-3/2
2)
P(4,-4) -->(-4, 7)
4 - 8 = -4 -------->left 8
-4 + 11 = 7 -------->up 11
Answer: left 8; up 11
3)
C(3,-1) , left 4 up 1
3 - 4 = -1 -------->left 4
-1 + 1 = 0 -------->up 1
a)
(x , y) -->(x - 4 , y +1)
C(3, -1) -->C'(-1 , 0)
b)
(x , y) --> (x - 4, y + 1); (-1 , 0)
Answer:
10365
Step-by-step explanation:
12579 - 2214
= 10365
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.