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.
A square would be X by X This rectangle is said to be (X-5) by (X+3)
(X-5)(X+3) = 240
(x+3)(x-5)=240?
Okay I got that
yes
What do I do now?
do you know how to find the area of a rectangle
Solve for X, you will get 2 answers, the positive one is correct
do I FOIL (x+3)(x-5)?
yes
I ended up getting x^2-2x-15
move the 240 over now to get x^2 - 2x - 255 = 0
factor that, or use the quadratic formula , since it is a book prob, it will probably factor nice
Do I make that into two binomials?
yep (x-17)(x+15)=0
Wait how did you get the 17 and 15?
255 = 17 * 15 and 15 - 17 = -2 (middle number)
If you multiply that out, you get back to the original ax^2 + bx + c = 0
Ohhh okay I understand that
or you could always just apply the quadratic formula, and you would get the X values
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
The problem is a physical measure, so the positive value is the one that makes sense... X=17
What do I write for part A and for Part B?
First Part... Area of Land = Length * Width = 240 240 = (X-5)(X+3)
And do I work it out until I get (x-17)(x+15) = 0?
right, for the second part you solve it
Thank you sososo much! You really helped a lot.
yw
Do you think you can check over some of my answers? If you have the time of course
sure
Thanks so much! They're quite long, so do you want me to screenshot it and attach it?
Sure, i have a bit of time
And that's it, since I don't want to have you check all of them, haha
The first one looks good, what is the prob with that one?
I just wanted to see if I got it correct or not
That one looks good to me
Thank you (:
second one looks good too
yay ^-^
For the last one part C, it asks if they all have a common factor
Yes, is it wrong?
I don't see a common factor btw all 3 of those
So it would be no because no common factor is consistent between the three polynomials?
yeah i think so
Do you think you could check 2 more?
k, real quick, then i have to go... you can fan me for help later if you want
Alright! Once I finish with these questions, I'll be done with Algebra for the year, aha
haha
And that's it!
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)
Okay
anything else?
nope, looks good to me
Yay!
Did you check the last one? or
Part b asks what the maximum profit is, you dont say what that is
How do I find the maximum?
it would be at the vertex of the parabola, or axis of symmetry value
How do I find that? Did my steps correspond to the question at all?
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)
Huh?
vertex point - (h,k) Vertex Equation - y = a(x-h)^2+k
Do I need to rewrite anything?
so the vertex from your equation is at the point, (5, 2250)
This is so confusing ._. I'm sorry that I'm so difficult to work with
sorry i said year... the most profit is at (5,2250) price 5 and prfit 2250
How do I find that with the steps I wrote down?
which is also on the axis of symmetry
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.
Ahhh okay! So now I write that the maximum profit is 2250 ?
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.
Thank you sosososo much!! You're truly a lifesaver, I wish I could give you a thousand medals , haha
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