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3 years ago

Simplify the function f

Mathematics

1 answer:

netineya [11]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is the coefficient of the x^5y^5 term in the binomial expansion of (2x-3y)^10?

nalin [4]

ANSWER

$10C_5(2)^{5}( - 3)^5$

EXPLANATION

The given binomial expression is

$(2x-3y)^{10}$

The nth term in this expansion is calculated using the formula,

$T_{r+1} = nC_ra^{n-r}b^r$

To find the term with coefficient

${x}^{5} {y}^{5}$

we substitute r=5, n=10, a=2x, and b=-3y

$T_{6} = 10C_5(2x)^{10-5}( - 3y)^5$

$T_{6} = 10C_5(2x)^{5}( - 3y)^5$

$T_{6} = 10C_5(2)^{5}( - 3)^5 {x}^{5} {y}^{5}$

The coefficient is

$10C_5(2)^{5}( - 3)^5$

The second choice is correct.

5 0

4 years ago

James is saving 5p and 10p coins. He has 75 coins. The ratio of the 5pto the 10p coins is 1 : 1.5. How much are his coins.

Ghella [55]

Answer:

the total ratio is 1+1.5=2.5

to get the number of 5p coins you get (1/2.5)*75=30

To get the number of 10p you get

(1.5/2.5)*75=45

so he has 30 5p coins and 45 10p coins so the total value Is given by

(5*30)+(10*45)=600

8 0

3 years ago

Suppose that a department contains 9 men and 15 women. How many different committees of 6 members are possible if the committee

Drupady [299]

Answer: The required number of different possible committees is 81172.

Step-by-step explanation: Given that a department contains 9 men and 15 women.

We are to find the number of different committees of 6 members that are possible if the committee must have strictly more women than men.

Since we need committees of 6 members, so the possible combinations are

(4 women, 2 men), (5 women, 1 men) and (6 women).

Therefore, the number of different committees of 6 members is given by

$n\\\\\\=^{15}C_4\times ^9C_2+^{15}C_5\times ^9C_1+^{15}C_6\\\\\\=\dfrac{15!}{4!(15-4)!}\times \dfrac{9!}{2!(9-2)!}+\dfrac{15!}{5!(15-5)!}\times \dfrac{9!}{1!(9-1)!}+\dfrac{15!}{6!(15-6)!}\\\\\\\\=\dfrac{15\times14\times13\times12\times11!}{4\times3\times2\times1\times11!}\times\dfrac{9\times8\times7!}{2\times1\times7!}+\dfrac{15\times14\times13\times12\times11\times10!}{5\times4\times3\times2\times1\times10!}\times\dfrac{9\times8!}{1\times8!}+\dfrac{15\times14\times13\times12\times11\times10\times9!}{6\times5\times4\times3\times2\times1\times9!}\\\\\\=1365\times36+3003\times9+5005\\\\=49140+27027+5005\\\\=81172.$