English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

please help!

1. the equation of a rectangle is 40 inches. find the dimensions of the rectangle if its area is 84 square inches.

2. find 2 consecutive integers whose product is 528.

PLEASE HELP!
thanks in advance

:]

2007-09-30 16:02:34 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

I believe your 'equation' of a rectangle is the perimeter. Then if L is length and W is width,
40=2L+2W or L=20-W. Since the area=LW, we have 84= W*(20-W), which expands out to
W^2 -20W +84 =0.
This has roots of 6 and 14. Either can be the width as long as the other is the length.

This problem doesn't work, since 529 is a square root. The two integers are probably the two consecutive EVEN integers 22 and 24.

2007-09-30 16:13:44 · answer #1 · answered by cattbarf 7 · 0 1

The trick with these sort of questions lies in developing your own quadratic to solve.

1. I assume you mean the PERIMETER of the rectangle is 40, as rectangles dont have an equation ;).

The perimeter of the rectangle is 2L +2B = 40

The area of a rectangle is LB = 84
therefore B = 84/L

If we substitute this into the perimeter equation:
2L + 2(84/L) = 40
2L^2 + 168 = 40L
L^2 -20L +84 = 0

Using the quadratic formula:
L = [20 (+/-) SQRT (400-4x84)]/2
=(20 (+/-) 8)/2
L = 14 or 6

Since B = 84/L
B = 6 or 14 - the same dimensions!

Therefore the rectangle is 6 inches by 14 inches.

2. Consecutive integers, product 528.
Let one integer be x. THerefore the other can be written as (x+1)

So:
x(x+1) = 528
x^2 + x -528 = 0
Back to the formula:

x = [-1 (+/-) SQRT (1-4 x -528)]/2
Unfortunately, this does NOT give a integer answer. Did you write the question down correctly?

2007-09-30 16:18:29 · answer #2 · answered by mevelyn2551 3 · 0 1

1. I'm going to assume that you meant the perimeter of a rectangle is 40 inches.

L = Length
W = Width

2L + 2W = 40
L*W = 84

W = 84/L

L + (84/L) = 20
L^2 + 84 = 20L
L^2 - 20L + 84 = 0
(L - 6)(L - 14) = 0

L = 6, L = 14

Meaning, L = 14, W = 6 or L = 6, W = 14

2. X(X+1) = 528
X^2 + x -528 = 0

There is not an integer solution for X, so there are no consecutive integers whose product is 528. Are you sure you wrote the problem down correctly?

2007-09-30 16:15:43 · answer #3 · answered by Will 3 · 0 0

1. Its A, The quadratic equation is A, quadratic equations are supposed to be in the this form - ax^2+bx+c 2. The value of coefficient a is 9 3. the vertex is B (1,-1) 4. Its opens UP and has a maximum point. Therefore C Hey, please rate this as best answer if it helped you, trynna level up you know. :)

2016-05-17 21:59:26 · answer #4 · answered by ? 3 · 0 0

Yeah... #2 is bunk.

2007-09-30 16:16:43 · answer #5 · answered by MacMooreno 2 · 0 1

fedest.com, questions and answers