Algebra2 word problem help?

The length of a rectangle is 12 meters longer than half the width.The area

of the rectangle is 90 square meters. Find the dimensions of the rectangle. Show work please, thanks.

Comments

  • l = 1/2w+12

    so

    (1/2w+12)w=90

    1/2w^2+12w=90

    then, multiply both sides by 2 to get rid of the 1/2.

    w^2+24w=180

    then subtract both sides by 180.

    w^2+24w-180=0

    (w-6)(w+30)=0

    w=-30,6

    You can't use a negative number in length, so the width is 6.

    1/2(6)+12 = 15

    So, the length is 15 and the width is 6.

    To check my work

    w*l=90

    6*15 = 90

    90 = 90

  • L = (1/2)W + 12 (length 12 more than half the width)

    L * W = 90 (area is 90)

    Substitute (1/2)W + 12 from equation 1 for L in equation 2:

    [(1/2)W + 12] * W = 90

    Multiply it out:

    (1/2)W^2 + 12W = 90

    Rewrite to quadratic form:

    (1/2)W^2 + 12W - 90 = 0

    Multiply through by 2 to clear the fraction:

    W^2 + 24W - 180 = 0

    Factor:

    (W - 6)(W + 30) = 0

    Discard the negative root:

    W = 6 meters

    L = 90/6 = 15 meters

  • Let w = width, l = length

    That first sentance in math is l = 12 + w/2.

    Area = l * w = 90

    Substituting for l: A = (12 + w/2)(w) = 90

    You should be able to solve from there.

    (Rearrange to get everything on one side, then use the quadratic formula to solve).

  • No on a million. Bro a million had x (unknown)money Bro 2 had x(unknown money)+$15 blended that they had $fifty 5 x+ (x+$15)=$fifty 5 (positioned like words at the same time ) 2x = -$15+ $fifty 5 2x=$40 x=$20 x+$15=$35 $20+$35=$fifty 5 as for mike and luke have been they in "X wing" combatants...290m/hr???

  • 90= w * (12)+(1/2)w is the equation.

Sign In or Register to comment.