So I'm guessing that the x2 is your way of saying 2 is the exponent right? In that case the equation would be 8x^2 + 12x
Well you would have to pull out the gcf which in this case will be 4x after pulling out the gcf you should get 4x(2x + 3)
-108=8x
Divide 8 on both sides
-13.5=x
The answer is...X=-13.5
PLEASE MARK ME AS THE BRAINLIEST
Answer:
Step-by-step explanation:
When given the following functions,
![g=[(-2,-7),(4,6),(6,-8),(7,4)]](https://tex.z-dn.net/?f=g%3D%5B%28-2%2C-7%29%2C%284%2C6%29%2C%286%2C-8%29%2C%287%2C4%29%5D)
One is asked to find the following,
1. Question 1
When finding the inverse of a function that is composed of defined points, one substitutes the input given into the function, then finds the output. After doing so, one must substitute the output into the function, and find its output. Thus, finding the inverse of the given input;
2. Question 2
Finding the inverse of a continuous function is essentially finding the opposite of the function. An easy trick to do so is to treat the evaluator (h(x)) like another variable. Solve the equation for (x) in terms of (h(x)). Then rewrite the equation in inverse function notation,
3. Question 3
This question essentially asks one to find the composition of the function. In essence, substitute function (h) into function (
) and simplify. Then substitute (-3) into the result.
Now substitute (-3) in place of (x),
Answer:
Two parallel lines
Step-by-step explanation:
the slopes are the same: slope = 3
f(x) 3x + 5 g(x) = 3x - 2
Pick at least three x values and find their corresponding values for y. Remember the f(x) and g(x) represent y
f(x) = 3x + 5 g(x) = 3x - 2
y = 3x + 5 y = 3x - 2
let x = 0 y = 3(0) + 5 y = 3(0) - 2
y = 5 y = -2
Ordered pair (0, 5) (0, -2)
Let x = 1 y = 3(1) +5 y = 3(1) - 2
y = 8 y = 1
ordered pair (1, 8) (1, 1)
Let x = -1 y = 3(-1) + 5 y = 3(-1) - 2
y = 2 y = -5
ordered pair (-1, 2) ( -1, -5)
Locate the ordered pairs and draw the graph on the lines.
you will graph two parallel lines
Answer:
I believe it is A!!!!!
Step-by-step explanation:
I took the test E2020
