Answer:
32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243
Step-by-step explanation:
(2x + 3)⁵
₅C₀ (2x)⁵ (3)⁰ + ₅C₁ (2x)⁴ (3)¹ + ₅C₂ (2x)³ (3)² + ₅C₃ (2x)² (3)³ + ₅C₄ (2x)¹ (3)⁴ + ₅C₅ (2x)⁰ (3)⁵
Use Pascal's triangle to find nCr.
1 (2x)⁵ (3)⁰ + 5 (2x)⁴ (3)¹ + 10 (2x)³ (3)² + 10 (2x)² (3)³ + 5 (2x)¹ (3)⁴ + 1 (2x)⁰ (3)⁵
(2x)⁵ + 15 (2x)⁴ + 90 (2x)³ + 270 (2x)² + 405 (2x)¹ + 243
32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243
Let the required point be (a,b)
The distance of (a,b) from (7,-2) is
=
But this distance needs to be betweem 50 & 60
So
Squaring all sides
2500 < (a-7)² + (b+2)² < 3600
Let a = 7
So we have
2500 < (b+2)² <3600
b+2 < 60 or b+2 > -60 => b <58 or b > -62
Also
b+2 >50 or b + 2 < -50 => b >48 or B < -52
Let us take one value of b < 58 say b = 50
So now we have the point as (7, 50)
The other point is (7,-2)
Distance between them
=
This is between 50 & 60
Hence one point which has a distance between 50 & 60 from the point (7,-2) is (7, 50)