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प्रश्न
The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
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उत्तर
Perimeter of the triangle = Sum of three sides
= 8y2 – 9y + 4
Sum of two sides = 3y2 – 5y + 4y2 + 12
= 7y2 − 5y + 12
∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)
= 8y2 – 9y + 4 – 7y2 + 5y – 12
= y2 – 4y – 8
Hence third side = y2 – 4y – 8
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संबंधित प्रश्न
Evaluate :
b2y − 9b2y + 2b2y − 5b2y
Evaluate :
abx − 15abx − 10abx + 32abx
Add : 13ab − 9cd − xy, 5xy, 15cd − 7ab, 6xy − 3cd
Add : x3 − x2y + 5xy2 + y3, - x3 − 9xy2 + y3, 3x2y + 9xy2
Subtract : 3a − 5b + c + 2d from 7a − 3b + c − 2d
Take m2 + m + 4 from −m2 + 3m + 6 and the result from m2 + m + 1.
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How much bigger is 5x2y2 – 18xy2 – 10x2y than –5x2 + 6x2y – 7xy?
