4. g(-2) = 3(-2), g(0) = 3(0), and g(5) = 3(5)
g(-3) = -6
g(0) = 0
g(5) = 15
5. h(-2) = -2(-2)+9, h(0) = -2(0) + 9, and h(5) = -2(5) + 9
h(-2) = 4 + 9 = 13
h(0) = 0 + 9 = 9
h(5) = -10 + 9 = -1
Essentially, all you're doing is plugging the x value into every single "x" you see in that function.
I'm pretty sure its anywhere between 2 and negative 2
Answer:
advance tickets $30.00
same day tickets $35.00
Step-by-step explanation:
1. In this equation, you are trying to find two values so you must use system of equations. Your two equations would be:
40x +20y = 1900
x + y = 65
x represents the price of advance tickets and y represents the price of same day tickets.
2. Solve the equation by performing elimination using multiplication. Multiple one equation by a constant to get two equations that contain opposite terms.
40x + 20y = 1900
-40(x + y = 65) → -40x + -40y = -2600
3. Add the equations, eliminating one variable. (add -40x + -40y = -2600 to 40x + 20y = 1900)
40x + 20y = 1900
+ -40x + -40y = -2600
Results in: -20y = -700
Solve for y by dviding both sides by -20.
y = 35
4. Substitute the solved value into one of the equations and solve for the remaining variable.
x + 35 = 65
Subtract 35 from both sides to solve for x.
x = 30
The advance tickets were $30.00 and the same day tickets were $35.00.
Answer:
x = -2
Step-by-step explanation:
( 2x - 1 ) ( x + 3 ) - ( x + 3 ) ( x - 3 ) = x + 2
Expand
x^2 + 5x + 6 = x + 2
x^2 + 5x + 6 - 2 = x + 2 - 2
x^2 + 5x +4 = x
x^2 + 5x + 4 - x = x - x
x^2 + 5x + 4 = 0
-4 / 2 = -2
x = -2
D. 24.84KM/H
Because 6.9m/s x 1KM/1000m x 3600s/1h = 24.84KM/H
