23 replies to this topic

[Frank12]

Ok, to start off, Im pretty sure this doesnt violate rules of any kind. If it does, ill just edit the board out or w/e. Regardless, its actually my brothers HW, tht i lost a bet on and now i have to do. Im willing to pay 500k for answer and brief explanation. I would do it myself, but i truthfully dont feel like doing the work out, and i have a mock trial i have to prep for, so i dont want to waste time. I cant guarantee NP to anyone except the person who PMs me first with the File or answers in the PM. I will screenshot to those who may object. Pretty much what im saying is there will be only one person who gets paid, and that is whoever PMs me first with the correct answers and brief explanation. I will pay SOMEONE, unless everyone gets them wrong. Its Algebra 2 tho, so it should just be an easy 500k.
Heres the problems....

1. You are playing a lawn version of TIC-TAC-TOE in which you toss bean bags onto a large board. One of your tosses can be modeled by the function ( y = -0.12x^2 + 1.2 x + 2) where x is the beanbags horizontal position (in feet) and y is the corresponding height (in feet). WHat is the beanbags maximum height?

2.The ball leaves your hand 5 feet above the ground and has an initial vertical velocity of 50 fps. You friend catches it when it falls to 3 feet. About how long is it in the air?

3.the path of a football kicked from the ground for a field goal try can be modeled by y = -0.013x(^2) + 0.67x where x is the horizontal distance in yards from where the ball was kicked and y is the corresponding height in yards.

a) a football cross bar is ten feet high. write and solve an inequality to find at what values of x the ball is high enough to go over the cross bar.

b) a football player kicks the ball toward the goal from a distance of 45 yards away. will the player make a field goal? explain.

4.A music store sells about 50 of a new drum per month at a price of $120 each. For each 5$ decrease in price, about 4 more drums per month are sold.
a) write a function that models the stores revenue from sales of the new drum model.
b) Write an inequality you can use to find the prices that result in a revenue of over 6500$
c)Solve the inequality from part (b) algebraically and graphically

5. You are designing a rectangular garden with tomatoes along the gardens top and right side. The tomatoes will take up 1/4 of the area of the garden. The garden measures 12 x 20 feet. What will the width x of the tomato area be? Explain why you must reject 1 solution.

Now i need all the answers written out to an extent, as well as explained. No shotty explanations, or i will send it back. Please do everything algebraically, i know how to do it by calculus, but it needs to be explained by algebra.

BTW, please PM!

please help

Edited by Frank12, 17 February 2011 - 05:04 PM.

[Noitidart]

Oh gosh lol. That desperate?

Its not against the rules. But I'm sure we can all help you. I have to get some stuff done then I'll try to take a look for free if I have some time.

For first one im sure that is an upside down parabola, so the max height the parabolas maximum. So if x = infiniti. Infiniti square times -infiniti is -infiniti. -infiniti +1.2*infiniti = 0. so you have 0 +2. so i think the max height is 2.

[Frank12]

Thanks Noit. I feel like an idiot, i know how to do the calculus way, but unfortunatly i need to explain how i came to those answers algebraically, based on the height equation and such, which for some reason, im pulling a brain dead on

[Noitidart]

Check out my edit, i think i solved problem 1.

Oh btw i just graphed the function here:
http://www.math.uri....ashmo/fungraph/

And im wrong. the max is 5.
gosh i forgot how to get max'es >_<

[Frank12]

It makes sense, but i think the teacher is looking for something along the lines of a step by step process of simplifying

Edit: Like i said before, i can find the answers in a relatively short amount of time, but explaining the algebra is my problem

Edited by Frank12, 17 February 2011 - 10:17 PM.

[Noitidart]

Oh ok so you have to complete the square on that equation to get in into standard form. Then you can find the max. I think. lol let me try

[Frank12]

completing the square...(b/2)^2? tht thing? DAYYYYUUUUMMM its been a while

[Noitidart]

haha ok it goes like this:
y = -.12x^2 + 1.2x + 2

we need one in front of x^2 so factor out -.12 from the Ax^2 + Bx so we get:
y = -.12(x^2 - 10x) + 2

now that 10 of the 10x lets divide that by 2 and then square it. so you get 25. so now add in 25 into the parantheses.
y = -.12(x^2 - 10x + 25) + 2

we added 25 into that so we need to add 25 * -.12 to that side of the eq to make it balance out. so 25 times -.12 is 3 so we get:
y = -.12(x^2 - 10x + 25) + 2 + 3

and (x^2 - 10x + 25) simplifies to (x-5)^2

so in standard form:
y = -.12(x-5^2) + 5

and 5 is the maxium/min of the parabola. in this case it is the max because it is an upside down from the -.12 in front of that term we see there.

yay there we go done with prob 1.

[Jennysaur]

I did #1-4 and pmed


@Noit for #1 you can just use -b/2a which is like wayyyy easier.

Edited by Jenniferrr, 17 February 2011 - 10:35 PM.

[Noitidart]

Oh I forgot that method.

How do you do that?
Plug -b/2a as your x value?



For #2 jen what did you get? what i did was set 3 = that equation + 5 then solved it and got 12.64ft. And then I just divied that by 50fps getting .2528 seconds. Is that right?

[Jennysaur]

Oh I forgot that method.

How do you do that?
Plug -b/2a as your x value?



For #2 jen what did you get? what i did was set 3 = that equation + 5 then solved it and got 12.64ft. And then I just divied that by 50fps getting .2528 seconds. Is that right?

x = -b/2a, find x, plug back into equation

for #2, use h=-16t^2+(vinitial)t+(hinitial)
its an equation you have to memorize/know

I took alg2 last year so lol fresh in my mind
I'm young (x

[Frank12]

Ok i still need #5. anyone?

[Noitidart]

Ohhh i see. Ok physics yuck.

Ok i think i got #5.

Its rectangle 12 by 20. So it looks like this attached pic 1.

Now we want top and right tomatoes with width of x so it should look like pic 2.

So 1/4 of the total area (12*20=240) is 240 / 4 = 60.

So now the area of the line part of pic 2 is the area of the green (12x) plus the area of the red (20x) minus the area of the square (x*x) times 1. because the square area gets counted twice in 12x and 20x area calcs so subtract one of the squares. so equation is:
area of tomato area = 12x + 20x - x^2
which = 32x - x^2

so now that equals 60. so your equation is:

60 = 32x - x^2

rearrange that to:
-x^2 + 32x + 60 = 0
and use quadratic formula to solve.
you get x = 2 and x = 30.

so your answer is x=2. because the width of the tomato area as 30 is not possible, the max dimensions of the garden are 12 one way and 20 the other way.

Attached Files

•  num 5.GIF   2.72KB   4 downloads

[Frank12]

Damn noit. Nice work. I owe u, just name it, and if its within my power, ill get it

[Noitidart]

haha no problem thanks!
post what jen sent you here i want to see that. man its been years since i took alg 2 it was fun!
oh and i made a nicer red/green box image man i love paint

Attached Files

•  num 5.GIF   2.74KB   1 downloads

[Frank12]

1. x=-b/2a
x=-1.2/(2*-.12)
x=5
Plug back into equation
y=5feet is maximum height

2. h=-16t^2+(vinitial)t+(hinitial)
3=-16t^2+50t+5
0=-16t^2+50t+2
t is about 3.16seconds

3. a) 10<-.013x^2+.67x
b) No, when you plug in, right side of eq is only 3.825

4.a) revenue = (50+4x)(120-5x)
b) 6500 < (50+4x)(120-5x)
c) 6500<6000+230x-20x^2
-20x^2+230x-500<0
x= about 2.91 or 8.59
120-5(2.91) = $105.45
or
120-5(8.59) = $77.05

noit, im kinda lost. Like i understand how to do these, but it seems like theres too much jumping around.

Edited by Frank12, 17 February 2011 - 11:23 PM.

[HeavenNEarth]

Ohhh i see. Ok physics yuck.

Ok i think i got #5.

Its rectangle 12 by 20. So it looks like this attached pic 1.

Now we want top and right tomatoes with width of x so it should look like pic 2.

So 1/4 of the total area (12*20=240) is 240 / 4 = 60.

So now the area of the line part of pic 2 is the area of the green (12x) plus the area of the red (20x) minus the area of the square (x*x) times 1. because the square area gets counted twice in 12x and 20x area calcs so subtract one of the squares. so equation is:
area of tomato area = 12x + 20x - x^2
which = 32x - x^2

so now that equals 60. so your equation is:

60 = 32x - x^2

rearrange that to:
-x^2 + 32x + 60 = 0
and use quadratic formula to solve.
you get x = 2 and x = 30.

so your answer is x=2. because the width of the tomato area as 30 is not possible, the max dimensions of the garden are 12 one way and 20 the other way.

I was too late D:

Anyway I'm just gonna post what I did , it was pretty fun doing these algebra stuff again ~ reminds me of old times

20*12=240 - Whole Garden

Stated that the tomatoes will take 1/4 of the space
So:
Free space = 3/4*240 = 180
Tomato space = 1/4*240 = 60

(1)Area of free space = (20-x)(12-x) = 180
(2)Area of Tomato Space = 12(x)+20(x) = 60

From - (1)

(20-x)(12-x)=180
240-20x-12x+x^2=180
240-32x+x^2=180
60-32x+x^2=0
(x-2)(x-30)=0

Thus: x-30=0 OR x-2=0
x=30(Reject) x=2

Reason
30cm is too large as the garden is only 12cm x 20cm

[Noitidart]

Haha yeah that was a fun problem. Thanks for sharing heaven.

[Las3rs]

For these kinds of problems, graphing calculators will do pretty much all the work for you. For example, problem #1 can simply be plugged into the calculator, graph it, and find the vertex.

[Frank12]

easier said when u have a graphing calc.

mine broke recently

anyways, how do i solve 3? i need to solve a. as well

[HeavenNEarth]

Haha yeah that was a fun problem. Thanks for sharing heaven.

Yea, it was prolly the only one I could do too ~ haha

But I got 1 thing i haev to clear up, I thought the acceleration is fixed at 10m/s ?

[Las3rs]

easier said when u have a graphing calc.

mine broke recently

anyways, how do i solve 3? i need to solve a. as well

You can find them online for free as well, although they'll have less features.

[Sweeney]

You seriously do maths in feet and yards and such?
How peculiar.

[jcrdude]

You seriously do maths in feet and yards and such?
How peculiar.

Making it more difficult makes us better with fractions, perhaps?

That is the only vaguely justifiable reason I could think of off the top of my head.