Answer:
c. x=-2
Step-by-step explanation:
256=(1/4)^(3x+2)
㏒(1/4) (256)=3x+2
-4=3x+2
3x=-6
x=-2
Let u = x.lnx, , w= x and t = lnx; w' =1 ; t' = 1/x
f(x) = e^(x.lnx) ; f(u) = e^(u); f'(u) = u'.e^(u)
let' find the derivative u' of u
u = w.t
u'= w't + t'w; u' = lnx + x/x = lnx+1
u' = x+1 and f'(u) = ln(x+1).e^(xlnx)
finally the derivative of f(x) =ln(x+1).e^(x.lnx) + 2x
Answer:
D
Step-by-step explanation:
vertically stretched is on the outside of the |x| and has to be grater than 1
Answer:
The answer to your question is: D = non of the above
Step-by-step explanation:
Data
-4x - 4y = -5 (I)
x - 2y = 1 (II)
Process
Find the slopes of both lines
(I) -4x - 4y = -5
-4y = 4x + 5
y = 4/-4 x + 5/-4
y = - x + 5/4
(II) x - 2y = 1
- 2y = - x + 1
y = -1/-2 x + 1/-2
y = 1/2 x - 1/2
The slopes are different slope 1 = -1 and slope 2 = 1/2, so there is no relation between the lines.