Slope=y2-y1/x2-x1
Slope=1.2-0/2-0
Slope= 0.6
(x+1)(x-4)-(x-2) = 0
x²-4x+x-4-x+2 = 0
x²-4x-2 = 0
/\ = (-4)² - 4*1*(-2)
/\ = 16 + 8
/\ = 24
x = (-(-4)+/- \/24) / 2
x = (4+/-\/4*6)/2
x = (4+/-2\/6)/2
x = 2+/-\/6
x' = 2+\/6
x" = 2-\/6
Answer:
Part A: Options D and E
Part B: Disagree
Step-by-step explanation:
Part A:
From the graph attached,
Option A
The object is launched from the ground.
False.
Since, the object is launched from the 5 feet above the ground.
Option B
The flight is symmetric about the line t = 10
False.
Since, the flight is symmetric about the line t = 12
Option C
The object reaches a maximum height of 12 feet.
False.
At t = 12 seconds maximum height of the object = 19.4 feet
Option D
The object is flying in the air during the interval 0 < t < 25.93
True.
Option E
The object reaches its maximum height after 12 seconds of flight.
True.
After 12 seconds object is at the maximum height 19.4 feet.
Therefore, Options D and E are the correct options.
Part B:
From the table attached,
h(4) = 13
h(20) = 13
At t = 4 and 13 seconds height of the object is same.
h(4) = h(20)
But Timothy said h(4) < h(20)
Therefore, disagree with the statement of Timothy.
Just dividing the volumes of the pan, 189 and one bar, 3, suggests that at most she'll be able to cut 63 bars.
But you'll also have to prove that you can divide it in a regular pattern. And you can:
9 divides by 1
10.5 divides by 1.5
2 divides by 2.
So no incomplete bars!
