Reposting Math story problem

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eb8524

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Reposting Math story problem

Please explain and solve. Thank you.
A rain gutter is made from sheets of aluminum that are 20 inches wide. Determine the depth of the gutter that will allow a cross-sectional area of 13 square inches. Show that there are 2 different solutions to the problem. Round to the nearest tenth of an inch.
x(20+2x)= 13.

Posted 39 day ago

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Answers (9)

eb8524

To Amar Soni.
pattie48@comcast.net

Posted 37 day ago

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Amar Soni

Please send youe email I wll send you the solution as attachment.

Posted 37 day ago

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Amar Soni

PLEASE FIND THE SOLUTION ATTACHED

Posted 37 day ago

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matherapist

The equation x(20+2x)=13 is incorrect!
Given that the aluminum strip is 20" wide, and assuming the cross section is a rectangle we have the area of a rectangle is width x height.

Let x be the height of the cross section, then we need x" from each side of the strip which will leave
(20-2x) for the length thus x(20-2x) = 13
Multiplying and collecting terms we have 20x - 2x^2-13=0
Multiply both sides by -1 to make the equation easier to handle and get 2x^2 - 20x +13 = 0
solve by the quadratic formula x ={ -b(+/-) [b^2-4ac]}/2a

This gives - -20 +/- ([-20]^2-4(2)(13))^.5 all over 2 times 2

Solving we get [20+(296)^.5]/4 and [20-(296)^.5]/4 0r x = 9.3" or x=.7"

For x =9.3, the width of the gutter would be 20-2(9.3) or 1.4 1.4 x 9.3 = 13.02 . This shape would not be practical as it is too narrow.

for x = .7, the width is 18.6 .7*18.6 =13.02. If it doesn't rain too hard, this gutter woulfd catch the rain without overflowing.

Note that the width in the second case is twice the height in the first case.

In practice the most useful cross section would be one that maximizes the cross sectional area since there is no top, the hight and width should be 20/3 "" or about 6.7".

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math_fanatic

I believe I can help you with this problem IF you can tell me more about how these sheets of aluminum are constructed to form the rain gutter. From, the problem it is implied that there are two different ways this gutter could be constructed. Also explain how you came up with x(20 + 2x) = 13. Is x the depth of the gutter? If it is, then by this equation you are saying that the width of the cross-sectional area is 2 times the depth of the gutter plus 20 inches. How many sheets of aluminum are you using? If you're only using one, it's impossible for one side of the sectional area to be more than 20 inches since the aluminum sheets are only 20 inches wide.

If you feel like you have not had this question answered adequately from the answers below, please repost with the added details I mentioned above and I will answer the question for you.

If you have an image of a diagram that could show me the construction of the rain gutter, you can send it to me at autotheist@gmail.com

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MFB13

I am not sure how the problem was exactly set up by getting this quadratic equation you wrote at the end, but getting the solutions to this quadratic equation is easy since there are two real solutions to this quadratic equation. The setup of this problem which aparently gives a quadratic equation indicates that the trough or well cross sectional area is of a parabolic curve and not a circular curve. So if we start with your quadratic equation setup you wrote which is

x(20+2x) = 13
Now after multiplying out the terms and transposing,
the above equation can also be written as x^2 + 10x - 13/2 = 0

The two real solutions to the above quadratic equation are

x1 = - 9.3 inch and x2 = - 0.7 inch.

Since both solutions are less than zero, I am not sure how that translates into a "real world" design. Nevertheless, the two x values apear to be located negative to the origin of x.

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jayteacher

The problem does not tell us what x represents. However, since it is an area problem, I can assume that x represents a distance.

First the student must manipulate the problem into standard form. (I have no idea if the student knows what standard form means, but the complexity of the problem indicates he or she might. If standard form is unclear, I suggest you look it up in your textbook.)

Use the distributive property on the equation.
20x + 2x^2 = 13
Now, standard form.
2x^2 + 20x – 13 = 0

Second, the student must know how to use the quadratic equation. (I have no idea if the student knows what quadratic equation means or how to use it. However, the complexity of the problem suggests he or she might. If the quadratic equation is unclear, I suggest you look it up in your textbook.)

x = (– 20 ± √(20^2 – 4 (2) (– 13))) / 2 (2)
x = (– 20 ± √(400 + 104)) / 4
x = (– 20 ± √(504)) / 4
x = (– 20 ± 22.45) / 4
x = 2.45 / 4 and x = – 42.45 / 4 (However, since x is a distance, x cannot be negative. So, discard the negative answer. I assume that when the question is telling you to “Show that there are 2 different solutions to the problem.” it is referring to a positive and a negative answer. Actually, there is only one answer to the problem that makes sense, since x cannot be negative.)

x = 0.6 (Rounded to the nearest tenth.)

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eb8524

20-2x^2 = 13, 0=2x^2 - 20 + 13. That's all. I don't know what to do after that.

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Doctor Bob

Please meet me half way, good buddy.
What have you tried already?
What ideas do you have about solving this problems?
I want to ASSIST you; I do not want to do it FOR you. Otherwise, what will you learn from it?

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