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प्रश्न
The adjacent sides of a rectangle are x2 – 4xy + 7y2 and x3 – 5xy2. Find its area.
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उत्तर
Reqd. area = (x2 – 4xy + 7y2) (x3 – 5xy2)
= x2 (x3 – 5xy2) – 4xy (x3 – 5xy2) + 7y2 (x3 – 5xy2)
= x5 – 5x3y2 – 4x4y + 20x2y3 + 7x3y2 – 35xy4
= x5 + 2x3y2 – 4x4y + 20x2y3 – 35xy4
= (x5 – 4x4y + 2x3y2 + 20x2y3 – 35xy4) sq. unit.
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संबंधित प्रश्न
Multiply: 2y − 4y3 + 6y5 by y2 + y − 3
Simplify : (px – q) (px + q)
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Find the value of (3x3) × (-5xy2) × (2x2yz3) for x = 1, y = 2 and z = 3.
If x = 2 and y = 1; find the value of (−4x2y3) × (−5x2y5).
Evaluate: xz (x2 + y2) for x = 2, y = 1 and z= 1.
Evaluate: 2x(3x – 5) – 5(x – 2) – 18 for x = 2.
Multiply and then verify :
−3x2y2 and (x – 2y) for x = 1 and y = 2.
Multiply: 2x2 – 4x + 5 by x2 + 3x – 7
Multiply: (ab – 1) (3 – 2ab)
