Math help.

Hey guys, I'm in Algebra 2 right now and my teacher just assigned a worksheet worth 40 points and I'm stumped on every problem. It probably would help if I had a graphing calculator, but, I don't.
Thanks in advance if you help.

Problems:

***Full credit will only be given if justification is shown and the answer is rounded correctly. If you use technology, show a diagram of what you used.***


1. The function y = -16t² + 336 models the height y in feet of a stone t seconds after it is dropped from the edge of a vertical cliff. How long will it take the stone to hit the ground? Round to the nearest hundredth of a second.

Code:

y = -16t² + 336
0 = -16t² + 336
-336 = -16t²
21 = t²
t = 4.582

2. Draw parallelogram ABCD. BC is 16 inches longer than AB. Find the height of the parallelogram to the nearest thousandth if the area is 2231 and m∠ABC = 60°. Avoid rounding in the middle of the problem.
Pictures I used:

Code:

A = bh
A = 2231
b = x
sin60 = (√3)/2 (Look at third picture)
h = (√3)/2 * (x + 16)
2231 = (x) * ((√3)/2x + (√3)/8)
2231 = (√3)/2x² + (√3)/8x
I used a graphing calculator here
Go into your Y= screen
Y0 = (√3)/2x² + (√3)/8x
Y1 = 2231
Find where they intersect and that's x (I don't have a graphing calculator on me)
So h = 2231/(x + 16)

3. During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The temperature T in degrees Fahrenheit x minutes after the machine is put into operation is modeled by T = -0.005x² + 0.45x + 125. After how many minutes will the temperature reach 200 degrees?

Code:

Open up Y=
Y1= -0.005x² + 0.45x + 125
Y2= 200
Find intersection
They don't intersect
No solution

4. The tale shows a meteorologists's predicted temperatures for an April day in Washington D.C. starting at 8:00 AM.
Time - Predicted Temperature (°F)
(0)8 AM - 44.11
(2)10 AM - 53.85
(4)12 PM - 59.46
(6)2 PM - 60.93
(8)4 PM - 58.26
(10)6 PM - 51.46

a) Using the temperature at 2 PM as the high temperature and the temperature at 8:00 AM, write a quadratic function in vertex form to estimate the temperature at a given time. Round to the nearest thousandth.

Code:

Vertex form: y = a(x-h)² + k
44.11 = a(0-6)² + 60.93
-16.82 = 36a
a = -16.82/36

b) Predict the temperature at 11:00 AM using your function from a).

Code:

y = -16.82/36 * (3-6)² + 60.93
solve
w/e answer

c) Use quadratic regression to find a quadratic function that predicts the temperature at a given time. Round to the nearest thousandth.

Code:

y = Ax² + Bx + C * 3
x = time(0, 2, 4, 6, 8, 10) y = temperature
1. 44.11 = A(0)² + B(0) + C
C = 44.11
2. 53.85 = A(2)² + B(2) + 44.11
9.74 = 4A + 2B
4A = 9.74 - 2B
A = 9.74/4 - 1/2B
3. 59.46 = 4²(9.74/4 - 1/2B) + 4B + 44.11
15.35 = 38.96 - 8B + 4B
-23.61 = -4B
B = 5.9025
A = 2.435 - (1/2)5.9025
A = -0.11125
y = -0.11125x² + 5.9025x + 44.11

d) To the nearest tenth of a degree and to the nearest minute, use the function form from c) to determine the high temperature and at what time it occured.
Too lazy to do this.

e) Predict at what time the temperature was 46 degrees.
Too lazy to do this.

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Thank you video for all your help.

 

You don't need a graphing calculator for any of these. www.wolframalpha.com

1. Set y=0. Then solve for t
2. BC=AB+16, height=AB*sin(60 degrees), 2231= BC*height. 2231=(AB+16)*AB*sin(60 degrees) through substitution
3. Set 200=T, then solve for X
4. Y = Ax^2+Bx+C. You have 3 points for (x,y), so you can solve for A, B, and C

 

2. When you say 2231 = BC*h, wouldn't it be AB*h since AB is the base?
3. Doesn't really help since you still have x² and x.
4. Explain?

 

2. Depends how you draw the parallelogram, I set AB to be one of the diagonals

.......B_______C

A_______D


Like that

3. Did they not teach you the quadratic formula?
http://www.purplemath.com/modules/quads/qform01.gif

4. Let's imagine a graph where 8:00 AM represents x=0, and every additional x is an hour. So 2:00 PM represents x=6 for example. You have f(0)=44.11, f(2)=53.85, f(4)=59.46, and that's enough to solve this problem. Plug in 0 for x and 44.11 for y, plug in 2 for x and 53.85 for y, and plug in 4 for x and 59.46 for y. Then you have 3 equations and 3 variables

 

2. Oh, I normally do:
A-B
D-C
And that would make the most since if angle ABC was 60 degrees.

3. Yeah they did but I fail at life. Oh, just looked at it and alright.

4. Oh alright. Wasn't really thinking.

Thanks again.

EDIT:
Not sure what you're getting at with #4.
I get:
C = 44.11
9.74 = 4A + 2B
15.35 = 16A + 4B

 

Read my edit.

 

You're very close. Set 4A = 9.74-2B, then solve for A by dividing by 4. Then plug that answer into 15.35 = 16(A) +4B to solve

 

Oh wouldn't have thought of that lol.

Did you get
A = -0.51625
B = 5.9025

?

EDIT:
How would I set it to vertex form for a)?