It will be whichever equation you see that has the "c" = +3, in the format:
F(x) = y = ax^2 + bx + c, where a, b, and c are all integers, such that plugging a "0" into the x's will give:
y = a•0^2 + b•0 + 3 = 0 + 0 + 3 = 3
1)
g+4-3g=1+g
subtracting g from both sides
4-3g=1
subtracting 4 from both sides
-3g=-3
g=1
2)
-6a+3=-3(2a-1)
extending the right side
-6a+3=-6a+3
adding 6a to both sides
3=3
which is an identity
3)
0.5b+4=2(b+2)
extending the right side
0.5b+4=2b+4
subtracting 0.5 b from both sides
4=1.5b+4
subtracting 4 from both sides
0=1.5b
b=0
4)
8-(3+b)=b-9
subtracting 8 from both sides
-(3+b)=b-17
extending the left side
-3-b=b-17
adding b to both sides
-3=2b-17
adding 17 to both sides
14=2b
b=7
Answer:
8
Step-by-step explanation:
Answer: This is either a trick question or real, but round it to the nearest thousanth I guess.
Ln x + xe^y = 1
1/x + xf'(x)e^y + e^y = 0
f'(x) = (-e^y - 1/x)/xe^y
f'(1, 0) = (-e^0 - 1/1)/e^0 = (-1 - 1)/1 = -2
Let the required equation of tangent be y = mx + c, where y = 0, m = -2, x = 1
0 = -2(1) + c = -2 + c
c = 2
Therefore, required equation is y = -2x + 2
