Fan and Medal!!
The area of a rectangular piece of land is 240 square meters.
If the length of the land was 5 meters less and the width was 3 meters more, the shape of the land would be a square.
Part A: Write an equation to find the width (x) of the land. Show the steps of your work.
Part B: What is the width of the land in meters? Show the steps of your work.

- anonymous

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- DanJS

A square would be X by X
This rectangle is said to be (X-5) by (X+3)

- DanJS

(X-5)(X+3) = 240

- anonymous

(x+3)(x-5)=240?

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- anonymous

Okay I got that

- DanJS

yes

- anonymous

What do I do now?

- anonymous

do you know how to find the area of a rectangle

- DanJS

Solve for X, you will get 2 answers, the positive one is correct

- anonymous

do I FOIL (x+3)(x-5)?

- DanJS

yes

- anonymous

I ended up getting x^2-2x-15

- DanJS

move the 240 over now to get
x^2 - 2x - 255 = 0

- DanJS

factor that, or use the quadratic formula , since it is a book prob, it will probably factor nice

- anonymous

Do I make that into two binomials?

- DanJS

yep
(x-17)(x+15)=0

- anonymous

Wait how did you get the 17 and 15?

- DanJS

255 = 17 * 15
and
15 - 17 = -2 (middle number)

- DanJS

If you multiply that out, you get back to the original ax^2 + bx + c = 0

- anonymous

Ohhh okay I understand that

- DanJS

or you could always just apply the quadratic formula, and you would get the X values

- DanJS

so now you have
(x-17)(x+15)=0
which means either term can be zero, so
(x-17)=0 or (x+15)=0
X=17 or X=-15

- DanJS

The problem is a physical measure, so the positive value is the one that makes sense...
X=17

- anonymous

What do I write for part A and for Part B?

- DanJS

First Part...
Area of Land = Length * Width = 240
240 = (X-5)(X+3)

- anonymous

And do I work it out until I get (x-17)(x+15) = 0?

- DanJS

right, for the second part you solve it

- anonymous

Thank you sososo much! You really helped a lot.

- DanJS

yw

- anonymous

Do you think you can check over some of my answers? If you have the time of course

- DanJS

sure

- anonymous

Thanks so much!
They're quite long, so do you want me to screenshot it and attach it?

- DanJS

Sure, i have a bit of time

- anonymous

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- anonymous

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- anonymous

And that's it, since I don't want to have you check all of them, haha

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- DanJS

The first one looks good, what is the prob with that one?

- anonymous

I just wanted to see if I got it correct or not

- DanJS

That one looks good to me

- anonymous

Thank you (:

- DanJS

second one looks good too

- anonymous

yay ^-^

- DanJS

For the last one part C, it asks if they all have a common factor

- anonymous

Yes, is it wrong?

- DanJS

I don't see a common factor btw all 3 of those

- anonymous

So it would be no because no common factor is consistent between the three polynomials?

- DanJS

yeah i think so

- anonymous

Do you think you could check 2 more?

- DanJS

k, real quick, then i have to go... you can fan me for help later if you want

- anonymous

Alright! Once I finish with these questions, I'll be done with Algebra for the year, aha

- DanJS

haha

- anonymous

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- anonymous

And that's it!

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- DanJS

For the first, for the domain, be sure to call the variable 't' and not x, Since y is a function of t, in S(t)

- anonymous

Okay

- anonymous

anything else?

- DanJS

nope, looks good to me

- anonymous

Yay!

- anonymous

Did you check the last one? or

- DanJS

Part b asks what the maximum profit is, you dont say what that is

- anonymous

How do I find the maximum?

- DanJS

it would be at the vertex of the parabola, or axis of symmetry value

- anonymous

How do I find that? Did my steps correspond to the question at all?

- DanJS

It looks like you have P(n) written in vertex form in part b right... so you can use that to extract the vertex coordinate (h,k)

- anonymous

Huh?

- DanJS

vertex point - (h,k)
Vertex Equation - y = a(x-h)^2+k

- anonymous

Do I need to rewrite anything?

- DanJS

so the vertex from your equation is at the point, (5, 2250)

- anonymous

This is so confusing ._. I'm sorry that I'm so difficult to work with

- DanJS

sorry i said year... the most profit is at (5,2250) price 5 and prfit 2250

- anonymous

How do I find that with the steps I wrote down?

- DanJS

which is also on the axis of symmetry

- DanJS

The last thing you wrote, is the vertex equation for the parabola.
vertex point - (h,k)
Vertex Equation - y = a(x-h)^2+k
The maximum profit 'k' is at that vertex point, you can just read it from your equation.

- anonymous

Ahhh okay! So now I write that the maximum profit is 2250 ?

- DanJS

yes, The vertex is at point (5,2250)
When the tickets are sold at 5 dollars each, the profit is the most at 2250 dollars.

- anonymous

Thank you sosososo much!! You're truly a lifesaver, I wish I could give you a thousand medals , haha

- DanJS

|dw:1433203687682:dw|

- DanJS

welcome,

- anonymous

Take care!!

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