Answer:
x=2, y=4
Step-by-step explanation:
When using elimination, the objective is to elimate one variable. In this case, we see that "y" can easily be eliminated by adding the two equations together since you will get 3y + (-3y) which will eliminate y because the value would become 0, letting us solve for x.
Then we get
, because the y's will eliminated, and 8x+7x is 15x, and 28+2 is 30.
Then divide by 15 on both sides and you get x=2
If x=2, then we can substitute that value into any of the previous given equations and find the value of y.
8×2 +3y = 28
16+3y=28
3y=12
y=4
So the answer to your system of equations would be x=2, and y=4
You can substitute the answers we found to see that they satisfy the equation.
Hope this helped.
Answer:
40
Step-by-step explanation:
This is asking you to plug in the value of h(5) into g(x). First solve for h(5)
x^2-3 = h(x)
5^2-3= 22
h(5) = 22
h(5) is 22. So now, plug that into g(x).
g(h(5)) = 2(22)-4
44-4=40
(g.h)(5) is 40.
Answer:
The answer is A) a=-4
Step-by-step explanation:
Answer:
a. 2^(x-2) = g^(-1)(x)
b. A, B, D
Step-by-step explanation:
the phrasing attached in the image is flagged as inappropriate, so i will be replacing it with g(x) and its inverse with g^(-1)(x)
1. replace g(x) with y and solve for x
y = log₂(x) + 2
subtract 2 from both sides to isolate the x and its log
y - 2 = log₂(x)
this text is replaced by the second image -- it was marked as inappropriate
thus, 2^(y-2) = x
replace x with g^(-1)(x) and y with x
2^(x-2) = g^(-1)(x)
2. plug this in to points A, B, C, D, E, and F
A: (2,1)
plug 2 in for x
2^(2-2) = 2⁰ = 1 so this works
B: (4, 4)
2^(4-2) = 2²= 4 so this works
C: (9, 3)
2^(9-2) = 2⁷ = 128 ≠ 3 so this doesn't work
(5, 8)
2^(5-2) = 2³ = 8 so this works
E: (3, 5)
2^(3-2) = 2¹ = 2 ≠ 5 so this doesn't work
F: (8, 5)
2^(8-2) = 2⁶ = 64 ≠ 5 so this doesn't work
