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3 years ago

What is the binomial expansion of (2x-3)^5

Mathematics

2 answers:

faltersainse [42]3 years ago

5 0

Answer:

(2x - 3)⁵= 32x⁵ - 240x⁴ + 720x³ - 1080x² + 810x - 243

Step-by-step explanation:

We need to write the expansion of Binomial (2x - 3)⁵

Here general form of binomial expansion is:

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ⁿC₃aⁿ⁻³b³ + ... + ⁿCₙbⁿ

(2x - 3)⁵= ⁵C₀(2x)⁵ + ⁵C₁(2x)⁵⁻¹(-3) + ⁵C₂(2x)⁵⁻²(-3)² + ⁵C₃(2x)⁵⁻³(-3)³

+⁵C₄(2x)⁵⁻⁴(-3)⁴ + ⁵C₅(2x)⁵⁻⁵(-3)⁵

(2x - 3)⁵= (32x⁵) + 5(16x⁴)(-3) + 10(8x³)(-3)² + 10(4x²)(- 3)³ + 5(2x)(-3)⁴+(-3)⁵

(2x - 3)⁵= 32x⁵ - 240x⁴ + 720x³ - 1080x² + 810x - 243

That's the final answer.

omeli [17]3 years ago

3 0

Answer: $32x^5-240x^4+720x^3-1080x^2+810x-243$

Step-by-step explanation:

Binomial expansion of $(a+b)^n= ^nC_0 a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+....+^nC_na^0b^n$

In $(2x-3)^5$ , a= 2x and b= -3

Similarly, Binomial expansion of $(2x-3)^5$

$= ^5C_0 (2x)^5(-3)^0+^5C_1(2x)^{4}(-3)^1+^5C_2(2x)^{3}(-3)^2+^5C_3(2x)^{2}(-3)^3+^5C_4(2x)^{1}(-3)^4+^5C_5(2x)^{0}(-3)^5\\\\=(1)(32x^5)+(5)(16x^4)(-3)+(\dfrac{5!}{2!3!})(8x^3) (9)+(\dfrac{5!}{2!3!})(4x^2) (-27)+(5)(2x)(81)+(1)(-243)\\\\=32x^5-240x^4+720x^3-1080x^2+810x-243$

Hence, Binomial expansion of $(2x-3)^5$

$=32x^5-240x^4+720x^3-1080x^2+810x-243$

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3 years ago

Suppose theta(????) measures the minimum angle between a clock’s minute and hour hands in radians. What is theta′(????) at 4 o’c

makvit [3.9K]

Answer:

$\frac{\pi }{30}$ radians per minute.

Step-by-step explanation:

In order to solve the problem you can use the fact that the angle in radians of a circumference is 2π rad.

The clock can be seen as a circumference divided in 12 equal pieces (because of the hour divisions). Each portion is $\frac{1}{12}$

So, you have to calculate the angle between each consecutive hour (Let ∅ represent it). It can be calculated dividing the angle of the entire circumference by 12.

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