1
because
1^2=1
1^2×15=15
1×15=15
So 1^2*15=1×15
Equation looks like this
15x^2=15x
subtract 15x
15x^2+15x=0
factor
15x(x-1)=0
solutions are 0 or 1 but seeing it said non zero number it must be 1
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: A
Step-by-step explanation:
r = j + 3
We subtitube r to find j,
r = 9 j = 9 + 3 = 12
r = 15 j = 15 + 3 =18
r = 21 j = 21 + 3 = 24
And we found A as answes.
Answer:
f(-2) = 1
Step-by-step explanation:
f(-2) means find the y value when x = -2
Going to x=-2 and going up until we hit the blue line
f(-2) = 1
Answer:
(a) 0.4
(b) a = 3
Step-by-step explanation:
(a) The area under the curve from x=4 to x=6 is 0.2 units high and 2 units wide, so is 0.2·2 = 0.4. (The area of a rectangle is the product of length and width.)
(b) The area is 0.2 and the height of the curve is 0.2, so the width of the region of concern is 0.2/0.2 = 1. (Again, area = height·width, or width = area/height.) 1 unit from the left end is found at X=3, so a = 3.