58
N i g g e r s
5g times g
what is the domain of g(x) equals square root of x plus 1? √(x+1) ≥ 0 x+1≥0 x≥-1 Domain: [-1,∞)
For these type of calculations its just a straightforward substitutation. So for g=5 we put in 5 for g. So that's 11(5) = 55 if you evaulated the expression for g=7 it would be 11(7) = 77 etc... for all whole number values of g.
2 + g + 5 = g + 7
gf(5) = g(f(5)) = g(5+1) since f(x) = x+1 and then g(6) = 3*6 - 2 = 18 - 2 = 16
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
-2
(g + 5)(g + 2)
g (x) = -3x+5 -3x= 5 x= -5/3
27-g = 2g+12 -3g = -15 g = 5
2.845 Kg
g(x) = 5. So whatever f(x) or f(-1) is, g of that is going to be 5.
If "g" has a value then it would just be 5+(the value) Otherwise it would be 5+g G is a variable and therefor could be anything. You can not solve the equation further. The answer is simply 5+G. If you have a previous declaration that states that G=2, then 5+G would equal 7. In computer programming, equations like this are useful for repetitious calculations. For example, if G is the variable for a score in a game, we would start with G=0 and every time you get a point, the computer would calculate G=G+1. G, currently 0, plus 1, equals 1. If you get another point then G, currently 1, plus 1, equals 2. This can be repeated as needed without changing the equation.
G(x) = log(2x) + 2, obviously!
2 H2 (g) + O2 (g) => 2 H2O (g) + 483.6 kJ