Answer:

Step-by-step explanation:
<u>Equation of a Polynomial</u>
Given the roots x1, x2, and x3 of a cubic polynomial, the equation can be written as:

Where a is the leading coefficient.
We know the three roots of the polynomial -6, -3, and 1, thus:

Since the y-intercept of the polynomial is y=90 when x=0:
90=a(0+6)(0+3)(0-1)
90=a(6)(3)(-1)=-18a
Thus
a = 90/(-18) = -5
The polynomial is:

We must write it in standard form, so we have to multiply all of the factors as follows:





vi is going in the positive direction (up). (That's my choice). a (acceleration) is going in the minus direction (down). The directions could be reversed.
Givens
vi = 160 ft/s
vf = 0 (the rocket stops at the maximum height.)
a = - 9.81 m/s
t = ????
Remark
YOu have 4 parameters between the givens and what you want to solve. Only 1 equation will relate those 4. Always always list your givens with these problems so you can pick the right equation.
Equation
a = (vf - vi)/t
Solve
- 32 = (0 - 160)/t Multiply both sides by t
-32 * t = - 160 Divide by -32
t = - 160/-32
t = 5
You will also need to solve for the height to answer part B
t = 5
vi = 160 m/s
a = - 32
d = ???
d = vi*t + 1/2 a t^2
d = 160*5 + 1/2 * - 32 * 5^2
d = 800 - 400
d = 400 feet
Part B
You are at the maximum height. vi is 0 this time because you are starting to descend.
vi = 0
a = 32 m/s^2
d = 400 feet
t = ??
formula
d = vi*t + 1/2 a t^2
400 = 0 + 1/2 * 32 * t^2
400 = 16 * t^2
400/16 = t^2
t^2 = 25
t = 5 sec
The free fall takes the same amount of time to come down as it did to go up. Sort of an amazing result.
Answer:
W = 8.2 cm
Step-by-step explanation:
The perimeter formula is P = 2W + 2L and here the perimeter is 65.6 cm.
Thus, 65.6 cm = 2W + 2L. Substituting 3W for L, we get
65.6 cm = 2W + 2(3W) = 8W.
Solving for W by dividing both sides by 8, we get W = 65.6 cm / 8, or
W = 8.2 cm
11x+11y
11(x+y)
Explanation
Step 1
Let
Carlos earns == 11 per hour
x represents the number of hours he worked in May
the,
the amount he earned in Mayis

y represent the number of hours he worked in June.
the amount he earned in June was

Step 2
the amount of money he earned is May and June is the sum of the values

I hope this helps you
A parallelogram should have 2 sets of parallel lines. Let's find the slope of line PQ and RS to test.
PQ:
(4-2)/(1-(-3))
2/4
1/2
RS:
(2-0)/(3-1)
2/2
1
Because 1 does not equal 1/2 (the slopes are different) the lines are not parallel. Thus, the figure is not a parallelogram.