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Sagot :
Answer:
of triangle =12base×height
Area of triangle =2x2−11x+15
base =2x−5
Plugging in the given;
2x2−11x+15=12(2x−5)×height
2x2−11x+15=2x−52×height
2x2−11x+151=2x−52×height
2(2x2−11x+15)=(2x−5)×height
4x2−22x+30=(2x−5)×height
4x2−22x+302x−5=height
Resolving the quadratic equation;
(4x2−22x+30)
Simplifying;
4x
Answer:
2x-6
Step-by-step explanation:
height : 2(2x^2-11x+15)/(2x-5)
(4x^2-22x+30)/(2x-5)
Δ/4 = 121 - 120 = 1
x1 = (11+1)/4 = 3
x2 = (11-1)/ 4 = 5/2
4(x-5/2)(x-3)
(4x-10)(x-3)/(2x-5)
2(2x-5)(x-3)/(2x-5)
2x-6
Proof
(2x-6)(2x-5)/2
2(x-3)(2x-5)/2
2x^2-5x-6x+15
2x^2-11x + 15
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