undeadknight26
  • undeadknight26
Evan is making a table that will be created in the shape of the figure below. The table top is a triangle attached to a rectangle. To purchase the right amount of paint, he needs to know the area of the table top. He can only spend $10 on paint, which is enough to cover 150 ft2 of surface area. What is the maximum length of the base of the rectangle he can build?
Algebra
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SOLVED
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katieb
  • katieb
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undeadknight26
  • undeadknight26
i put it in wrong subject :p
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anonymous
  • anonymous
f*cking f@g no is going to help u
anonymous
  • anonymous
one*

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undeadknight26
  • undeadknight26
Why not?
goformit100
  • goformit100
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jdoe0001
  • jdoe0001
anyhow, the wording is a bit off, thus I had to check and recheck it turns out that Evan, is going to build the table, he knows the Area of the triangular part, but he doesn't know the Area of the rectangular part he can only use $10 for painting, after he's done that is, which will yield 150 square feet so whatever the Area of the rectangular part is, must fit within that constraint so the Area of the triangle, ADDED to the Area of the rectangle must be EQUALS OR LESS than 150 it can't be more than 150, because Evan is not allowed to spend for more than that for the paint when he's done \(\bf \textit{area of a triangle } = \cfrac{1}{2} (base)(height) \implies \cfrac{1}{2} (4)(6)\\ \textit{area of a rectangle} = x \times 6\\ \textit{triangle + rectangle areas} = \left[\cfrac{1}{2} (4)(6)\right] + \left[x \times 6\right]\\ \textit{thus our inequality}\\ \left[\cfrac{1}{2} (4)(6)\right] + \left[x \times 6\right] \le 150 \implies \cfrac{24}{2}+6x \le 150 \implies 12 + 6x \le 150\) I think you can get it from there :)
undeadknight26
  • undeadknight26
so i have to simplify?
undeadknight26
  • undeadknight26
So the answer is 132?
jdoe0001
  • jdoe0001
well, I didn't get 132 for "x", but yes, you have to simplify and solve for "x"
undeadknight26
  • undeadknight26
wouldn't u do 150-12 138-6? and then u have x left x=132?
jdoe0001
  • jdoe0001
ahemm. well, notice that 6 is MULTIPLIER, a factor, is not ADDING or SUBTRACTING in the inequality
undeadknight26
  • undeadknight26
so u would divide it not subtract!!! so 138/6!!!
undeadknight26
  • undeadknight26
so the answer would be 23?
jdoe0001
  • jdoe0001
\(\bf 12 + 6x \le 150 \implies 6x \le 138 \implies \cfrac{6x}{6} \le \cfrac{138}{6}\) yes, \(\bf x \le 23 \) :)
undeadknight26
  • undeadknight26
yay ty can u help me with 2 more?
undeadknight26
  • undeadknight26
they should not take that long
jdoe0001
  • jdoe0001
if I have the time, sure, post anew so we can all see it and help and revise each other :)
undeadknight26
  • undeadknight26
Im actually in the test right now ...
undeadknight26
  • undeadknight26
here is the question... As the number of pages in a photo book increases, the price of the book also increases. There is an additional shipping charge of 15%. The price of a book can be modeled by the equation below where, P = the price of the book, 20 is the printing charge, 0.5 is the charge per page, and x = the number of pages. P = (20 + 0.5x) + 0.15(20 + 0.5x) Jennifer wants to purchase a book but only has $62.10 to spend. What is the maximum number of pages she can have in her book? i got 273.33
jdoe0001
  • jdoe0001
hmm.. a bit too much
undeadknight26
  • undeadknight26
sorry
undeadknight26
  • undeadknight26
116.16?
jdoe0001
  • jdoe0001
\(\bf P = (20 + 0.5x) + 0.15(20 + 0.5x)\\ \textit{she would like to only spend }\$62.10\\ P = 62.10\\ \quad \\ 62.10 = (20 + 0.5x) + 0.15(20 + 0.5x)\\ \quad \\ \implies 62.10 = 20+0.5x + 0.15(20) + 0.15(0.5x)\)
undeadknight26
  • undeadknight26
23.65?
undeadknight26
  • undeadknight26
Hi joe im running out of time on my test...i have like 3 minutes left...
jdoe0001
  • jdoe0001
well, just add and divide terms
undeadknight26
  • undeadknight26
20.87?
jdoe0001
  • jdoe0001
she got more pages than that as far as helping with tests, well, you're supposed to cover all this before that, we can help you if you don't understand something, but the tests are there for you to do them
undeadknight26
  • undeadknight26
thanks for ur help doe i will try to relearn the formula!
jdoe0001
  • jdoe0001
yw
undeadknight26
  • undeadknight26
one more woule it be 41.23
anonymous
  • anonymous
I don't know

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