You have to derive for a multiplication in both terms:
=e^x+xe^x-(e^x-1 + (x-2)e^x-1) now apply distributive property in the last term:
=e^x+xe^x+e^x-1-xe^x-1 now replace each x by 0 (x=0)
=1 + 0 + e^-1 + 0 = 1+ e^-1 = 1.3679
What do you mean
Could you please insert an equation or a problem
15. 
Add "g" on both sides
Multiply 5 on both sides to get x by itself
x = 5(a + g)
x = 5a + 5g
18. a = 3n + 1
Subtract 1 on both sides
a - 1 = 3n
Divide 3 on both sides to get n by itself
= n
21. M = T - R
Add "R" on both sides to get "T" by itself
M + R = T
24. 5p + 9c = p
Subtract "5p" on both sides
9c = p - 5p
9c = -4p
Divide 9 on both sides to get "c" by itself
c = 
or c = 
27. 4y + 3x = 5
Subtract "4y" on both sides
3x = 5 - 4y
Divide 3 on both sides to get "x" by itself
x = 
x = 
Answer:
- 12 ft parallel to the river
- 6 ft perpendicular to the river
Step-by-step explanation:
The least fence is used when half the total fence is parallel to the river. That is, the shape of the rectangle is twice as long as it is wide.
72 = W(2W)
36 = W²
6 = W . . . . . . the width perpendicular to the river
12 = 2W . . . . the length parallel to the river
_____
<em>Development of this relation</em>
Let T represent the total length of the fence for some area A. Then if x is the length along the river, the width is y=(T-x)/2, and the area is ...
A = xy = x(T -x)/2
Note that the equation for area is that of a parabola with zeros at x=0 and at x=T. That is, for some fence length T, the area will be a maximum at the vertex of this parabola. That vertex is located halfway between the zeros, at ...
x = (0 +T)/2 = T/2
The corresponding area width (y) is ...
y = (T -T/2)/2 = T/4
Equivalently, the fence length T will be a minimum for some area A when x=T/2 and y=T/4. This is the result we used above.
Answer:
(x - 6) meters
Step-by-step explanation:
x² - 12x + 36
x² - 2(x)(6) + 6²
(x - 6)² = side²
side = x - 6
