SOLUTION: I need help. I am lost on this algebra.
The area of a rectangle is 119 m2. The base is four less than three times the height. What is the length of the base?
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-> SOLUTION: I need help. I am lost on this algebra.
The area of a rectangle is 119 m2. The base is four less than three times the height. What is the length of the base?
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Question 61017: I need help. I am lost on this algebra.
The area of a rectangle is 119 m2. The base is four less than three times the height. What is the length of the base?
You can put this solution on YOUR website! AREA=1/2BASE*HEIGHT
BASE=3*H-4
119=1/2(3*H-4)H
119=1/2(3H-4)H
119=1/2(3H^2-4H)
119=1.5H^2-2H
1.5H^2-2H-119=0 USING THE QUADRATIC EQUATION X=(-B+-SQRT[B^@-4AC])/2A WE GET
X=(2+-SQRT(4-4*1.5*-119])/2*1.5
X=(2+-SQRT[4+714])/3
X=(2+-SQRT[718])/3
X=(2+-26.8)/3
X=28.8/3
X=9.6 SOLUTION. THUS THE HEIGHT IS 9.6M
119=1/2(B*9.6)
119=9.6B/2
9.6B=2*119
9.6B=238
B=238/9.6
B=24.8 SOLUTION & THE LENGTH OF THE BASE.
PROOF
119=1/2(9.6*24.8)
119=1/2(238)
119=119
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