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प्रश्न
The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
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उत्तर
Required perimeter = Sum of three sides
= x2 − 3xy + 8 + 4x2 + 5xy − 3 + 6 − 3x2 + 4xy
= x2 + 4x2 − 3x2 − 3xy + 5xy + 4xy + 8 − 3 + 6
= 2x2 + 6xy + 11
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संबंधित प्रश्न
Evaluate :
3x2 + 5xy − 4y2 + x2 − 8xy − 5y2
Add : 5a + 3b, a − 2b, 3a + 5b
Add : x3 − x2y + 5xy2 + y3, - x3 − 9xy2 + y3, 3x2y + 9xy2
Subtract : 4xy2 from 3xy2
Subtract : −2x2y + 3xy2 from 8x2y
Take away – 3x3 + 4x2 – 5x+ 6 from 3x3 – 4x2 + 5x – 6
Subtract the sum of 5y2 + y – 3 and y2 – 3y + 7 from 6y2 + y – 2.
What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?
The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.
What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?
