PYQs for Class 9 Math Section B - Salwan Public School Gurgaon

Solution

∠1 = 55°

∠2 + ∠3 = 20° + 35° = 55°

∴ ∠1 = ∠2 + ∠3 = 55°

i.e. ∠ABM = ∠BMN = 55°

=> AB ∥ MN ( ∵ Alternate interior angles are equal lines are parallel ) - (1)

Also ∠3 + ∠4 = 35° + 145° = 180°

=> CD ∥ MN - (2)

From (1) and (2)

AB ∥ CD

Solution

4 (216)^-2/3    ₋      1 (256)^-3/4

4 {(6)³}^-2/3    ₋      1 (4⁴)^-3/4

4 6^-2    ₋      4 4^-3

= 4 x 6² - 4³

= 144 - 64

= 80

Solution

We know that (a - b)³ = a³ - b³ - 3ab(a - b)

= a³ - b³ - 3a²b + 3ab²

∴ (5x - 7y)³ = (5x)³ - (7y)³ - 3(5x)² x 7y + 3(5x)(7y)²

= 125x³ - 343y³ - 525x²y + 735xy²