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January 28, 2015

January 28, 2015

Posted by **Chirs** on Thursday, February 19, 2009 at 10:49pm.

1) m^-3m-2

2) 6c^-5c-4

3) ab+mn+an+mb

4) The length of a rectangular garden is 5ft more than the width. The garden has a 3ft wide cement walk sorrounding it. The total area of the garden and the walk is 546ft^. Find the dimensions of the garden

5) Find 2 consecutive integers such that the sum of their squares is 52

6) The perimeter of a rectangle is 22ft and the area is 24ft^. Find the dimensions of the rectangle

- Algebra -
**jane**, Thursday, February 19, 2009 at 11:42pm4)

GARDEN

width = w

length = w +5

GARDEN w/CEMENT WALK

width = w + 6

length = w + 5 + 6 = w + 11

a = 546 ft sq.

a = lw

a = (w+6)(w+11)

546 = (w+6)(w+11)

546 = w^2 + 17w + 66

0 = w^2 + 17w - 480

w = (-17 +/- sqrt(289 - 4(1)(-480))/2

w = (-17 +/- sqrt(289 + 1920))/2

w = (-17 +/- sqrt(2209))/2

w = (-17 +/- 47)/2

has to be + because it must be positive

w = 30/2 = 15

l = w + 5 = 20

- Algebra -
**jane**, Thursday, February 19, 2009 at 11:46pm6)

P = 22

A = 24

2l + 2w = 22

lw = 24

l = 24/w

2(24/w) + 2w = 22

48/w + 2w = 22

48 + 2w^2 = 22w

w^2 - 11w + 24 = 0

(w - 8)(w - 3) = 0

w = 8 or 3

l = 3 or 8

dimensions = 3 x 8

- Algebra -
**jane**, Thursday, February 19, 2009 at 11:47pm3) ab+mn+an+mb

ab + an + mn + mb

a(b + n) + m(n+b)

(a + m)(b + n)

- Algebra -
**drwls**, Friday, February 20, 2009 at 12:05am5) No solution. Obviously 4 and 5 give a sum of squares too low (41) and 5 and 6 give a sum of squares that is too high (61).

If you had said two consecutive EVEN integers, the answer would be 4 and 6.

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