Answer:
Step-by-step explanation:
We are given the function:
Let's find the inverse of g.
Call y=g(x):
We need to solve for x. Multiply both sides by x-2 to eliminate denominators:
Operate:
Collect the x's to the left side and the rest to the right side of the equation:
Factor the left side and operate on the right side:
Solve for x:
Interchange variables:
Call y as the inverse function:
Answer:
28
Step-by-step explanation:
cuz if u look at it you will see it adding by 4 each time, so 24+4=28
Hope I helped you out :)
Answer:
1. x = 2
2. x = 61/25
Step-by-step explanation:
Solve for x:
5 (x - 2) - 3 (2 - x) = 0
-3 (2 - x) = 3 x - 6:
3 x - 6 + 5 (x - 2) = 0
5 (x - 2) = 5 x - 10:
5 x - 10 + 3 x - 6 = 0
Grouping like terms, 5 x + 3 x - 10 - 6 = (3 x + 5 x) + (-6 - 10):
(3 x + 5 x) + (-6 - 10) = 0
3 x + 5 x = 8 x:
8 x + (-6 - 10) = 0
-6 - 10 = -16:
8 x + -16 = 0
Add 16 to both sides:
8 x + (16 - 16) = 16
16 - 16 = 0:
8 x = 16
Divide both sides of 8 x = 16 by 8:
(8 x)/8 = 16/8
8/8 = 1:
x = 16/8
The gcd of 16 and 8 is 8, so 16/8 = (8×2)/(8×1) = 8/8×2 = 2:
Answer: x = 2
_____________________________
Solve for x:
Solve for x:
3 (2 x - 7) + (7 x + 2)/3 = 0
Put each term in 3 (2 x - 7) + (7 x + 2)/3 over the common denominator 3: 3 (2 x - 7) + (7 x + 2)/3 = (9 (2 x - 7))/3 + (7 x + 2)/3:
(9 (2 x - 7))/3 + (7 x + 2)/3 = 0
(9 (2 x - 7))/3 + (7 x + 2)/3 = (9 (2 x - 7) + (7 x + 2))/3:
(9 (2 x - 7) + 2 + 7 x)/3 = 0
9 (2 x - 7) = 18 x - 63:
(18 x - 63 + 7 x + 2)/3 = 0
Grouping like terms, 18 x + 7 x - 63 + 2 = (18 x + 7 x) + (2 - 63):
((18 x + 7 x) + (2 - 63))/3 = 0
18 x + 7 x = 25 x:
(25 x + (2 - 63))/3 = 0
2 - 63 = -61:
(25 x + -61)/3 = 0
Multiply both sides of (25 x - 61)/3 = 0 by 3:
(3 (25 x - 61))/3 = 3×0
(3 (25 x - 61))/3 = 3/3×(25 x - 61) = 25 x - 61:
25 x - 61 = 3×0
0×3 = 0:
25 x - 61 = 0
Add 61 to both sides:
25 x + (61 - 61) = 61
61 - 61 = 0:
25 x = 61
Divide both sides of 25 x = 61 by 25:
(25 x)/25 = 61/25
25/25 = 1:
Answer: x = 61/25
So, notice, the focus point is at -7, 5, and the directrix is at y = -11.
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-7\\ k=-3\\ p=8 \end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2 \\\\\\ 32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2 \\\\\\ y=\cfrac{1}{32}(x+7)^2-3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-7%5C%5C%0Ak%3D-3%5C%5C%0Ap%3D8%0A%5Cend%7Bcases%7D%5Cimplies%204%288%29%5By-%28-3%29%5D%3D%5Bx-%28-7%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A32%28y%2B3%29%3D%28x%2B7%29%5E2%5Cimplies%20y%2B3%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2-3)
X = 5 and 2x = 10
the only solution is 5, can be any number
