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:
When we have a point (x, y) and we do a reflection over a given line, we know that the new point (x', y') will be at the same distance from the line as our initial point (x,y).
Now, in this case, we have a reflection over the line y = -1. (this line is parallel to the x-axis)
But in the image, we can see that the reflected triangle is drawn in the other side of the y-axis, this means that the reflection was made in a line parallel to the y-axis.
Then the mistake that Oscar did is that he reflected over the wrong line, seems that he reflected the triangle over the line x = -1 instead of the line y = -1.
Use FOIL
<span>(1-p)(1+p+p^2+p^3+p^4+p^5+p^6)
</span><span>1+p+p^2+p^3+p^4+p^5+p^6 -1p - p^2 - p^3 - p^4 - p^5 - p^6
</span><span>1 + p -1p <------------------Answer</span>
Divide 18 by 3 and double it to get 12 :))))))
Area = length*width
.. = (1*10^5 mm)*(8*10^4 mm)
.. = (1*8)*10^(5+4) mm^2
.. = 8*10^9 mm^2
Selection A is appropriate.
