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Algebra

posted by on .

Were doing factoring and i don't get these (^ means squared)
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 - ,

    4)
    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 - ,

    6)

    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 - ,

    3) ab+mn+an+mb
    ab + an + mn + mb
    a(b + n) + m(n+b)
    (a + m)(b + n)

  • Algebra - ,

    5) 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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