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???
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.