find the perimeter of triangle whose sides are 2a²+3a+2, 3a²-4a-1, 6a²+2y-2
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Answer: Perimeter of triangle= Sum of 3 sides
2a²+3a+2+ 3a²-4a-1+ 6a²+2y-2= 11a²-a+2y-1
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Step-by-step explanation:
Perimeter of a Triangle = a+b+c
A = 2a² + 3a + 2
B = 3a² - 4a - 1
C = 6a² + 2a - 2
A+B+C = (2a² + 3a + 2) + (3a² - 4a - 1) + (6a² + 2a - 2)
11a² + a - 1
b² - 4ac
1² - 4 (11)(-1) = 45
-b ± √b²-4ac
2à
-1 ± √45
22
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