Advertisements
Advertisements
प्रश्न
The adjacent sides of a rectangle are x2 – 4xy + 7y2 and x3 – 5xy2. Find its area.
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Multiply: 6x3 − 5x + 10 by 4 − 3x2
Simplify : (7x – 8) (3x + 2)
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.
Evaluate (x5) × (3x2) × (-2x) for x = 1.
If x = 2 and y = 1; find the value of (−4x2y3) × (−5x2y5).
Evaluate: (3x – 2)(x + 5) for x = 2.
Evaluate: xy2(x – 5y) + 1 for x = 2 and y = 1.
Multiply and then verify :
−3x2y2 and (x – 2y) for x = 1 and y = 2.
Simplify : (5 – x) (6 – 5x) (2 -x).
