Answer:
Part A: 2k(2c2+5(2)-8c-20)
Part B: sorry i don't know how to do this part
Step-by-step explanation:
Part A, you divide each number by 2 first so you get a simplified version of each number. Then, you will quickly realize that the varible k is similar in all of the numbers then you remove that and put it with the 2 outside the (). Hope you understood my explination.
Answer: 1.4 seconds
<u>Step-by-step explanation:</u>
The equation is: h(t) = at² + v₀t + h₀ where
- a is the acceleration (in this case it is gravity)
- v₀ is the initial velocity
- h₀ is the initial height
Given:
- a = -9.81 (if it wasn't given in your textbook, you can look it up)
- v₀ = 12
- h₀ = 3
Since we are trying to find out when it lands on the ground, h(t) = 0
EQUATION: 0 = 9.81t² + 12t + 3
Use the quadratic equation to find the x-intercepts
a=-9.81, b=12, c=3

Note: Negative time (-0.2) is not valid
What are the questions on 7 and 8?
It should be the 2nd or 4th answer, i’m not COMPLETELY sure. I’m positive it’s the 4th answer
Y-5=3-9 (y+2)
Solve for y
Distribute the 9 to (y+2)
Y-5=3-9y-18
Y-5=-15-9y
+9y to both sides
10y-5=-15
+5 to both sides
10y=-10
÷10 both sides
Y= -1
2 (x-7)-10=12-4x
Solve for X
Distribute 2 to (x-7)
2x-14-10=12-4x
2x-24=12-4x
+4x to both sides
6x-24=12
+24 to both sides
6x=36
÷6 to both sides
X=6