Answer:
3 ln(x − 1) − 3/(x − 1) + C
Step-by-step explanation:
∫ 3x / (x − 1)² dx
3 ∫ x / (x − 1)² dx
If u = x − 1, then x = u + 1, and du = dx.
3 ∫ (u + 1) / u² du
3 ∫ (1/u + 1/u²) du
3 (ln u − 1/u + C)
3 ln u − 3/u + C
Substitute back:
3 ln(x − 1) − 3/(x − 1) + C
F(x) + k - Moves the graph k units up.
k f(x) stretches the graph parallel to y-axis by a facor k
f (kx) stretches the graph by a factor 1/k parallel to x-axis
f(x + k) moves the graph 3 units to the left.
For k negative the first one moves it k units down
for second transform negative does same transfoormation but also reflects the graph in the x axis
For the third transform negative k :- same as above but also reflects in y axis
4th transform - negative k moves graph k units to the right
42 I think. I'm sorry if I'm wrong.. Hope this helped!!
Answer:
A. EG = √3 × FG
D. EG = √3/2 × EF
E. EF = 2 × FG
Step-by-step explanation:
∵ tan 60 = √3
∵ tan60 = EG/GF
∴ EG/GF = √3
∴ EG = √3 × GF ⇒ A
∵ m∠F = 60°
∵ sin60 = √3/2
∵ sin 60 = EG/EF
∴ √3/2 = EG/EF
∴ EG = √3/2 × EF ⇒ D
∵ cos60 = 1/2
∵ cos60 = GF/EF
∴ GF/EF = 1/2
∴ EF = 2 × GF ⇒ E
8y - 6 = 5y + 12
- 5y - 5y
3y - 6 = 12
+ 6 + 6
3y = 18
3 3
y = 6
