Answer:I have this and I’m in 7th grade lol
Step-by-step explanation:
Y=2x-4/3
x=2y-4/3 Switch variables
x/2 + 2/3 = y Isolate/Solve for y
H(x)^-1 = x/2 +2/3 Mark new equation as the inverse of the original.
There is an error in the second step. Before cross multiplying the denominators, LCM of the values should be taken.
Step-by-step explanation:
Step 1; 1 -
=
.
By taking an LCM of x-2 on the LHS, we get
-
=
,
as both terms on the LHS have same denominator, we can add them up,
=
=
.
Step 2; Now we cross multiply the denominators.
(x - 4) × (x + 2) = (x + 1) × (x - 2),
x² - 2x - 8 = x² - x - 2,
(x² - 2x - 8) - (x² - x - 2) = 0.
x² - x² - 2x + x - 8 + 2 = 0,
-x -6 = 0,
x = -6.
By simplifying the given expression, we get x = -6.
Any number to the 0 exponent is equal 1, except 0
x = 0, f(x) = 2(0.5)^0 = 2 x 1 = 2
Only the fist table has x = 0 and f(x) = 2
Plug in the rest, matching f(x) in the table 1
x = 1, f(x) = 2(0.5)^1 = 2 x 0.5 = 1
x = 2, f(x) = 2(0.5)^2 = 2 x 0.25 = 0.5
x = 3, f(x) = 2(0.5)^3 = 2 x 0.125 = 0.25
Answer is A. first table