Write a subtraction expression that involves two negative integers and has a positive answer.

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

One sandwich is shared equally among eight people. How much will each person get

givi [52]

Answer:

1/8 of a sandwich

3 0

2 years ago

Read 2 more answers

Help, Please! I really need this done right Carlos is analyzing the results of an experiment on two groups of mice in which he c

Allushta [10]

Given:
difference in the mean weight gain is 0.60 grams
standard deviation of the difference in sample mean is 0.305

68% confidence interval for the population mean difference is a) 0.305

0.60 + 1 * 0.305
0.60 + 0.305 = 0.905
0.60 - 0.305 = 0.295

95% confidence interval for the population mean difference is c) 0.61

0.60 + 2 * 0.305
0.60 + 0.61 = 1.21
0.60 - 0.61 = -0.01

7 0

3 years ago

Solve the system by substitution. Check your solution.

zvonat [6]

B. (9,126)

y + 18 = 16x
=>y=16x-18
0.5x + 0.25y = 36 (multiply both sides by 4)
=>2x+y = 144
Substitute y=16x-8

=>2x+16x-8=144
=>18x=152
=>x=152/18=9
y=16x-18
=>y=16(9)-18
=>y=144-18=126
Answer: x=9 and y=126

4 0

2 years ago

I don't really understand the question so can someone help me answer the question please?

Ksivusya [100]

Answer:

1/ 3

Step-by-step explanation:

If each of the cards is turned over, the probability of picking up a card of one type P(E) becomes equal to:

=> P(E) = number of cards of the required type/ total number of cards

● Total number of spades( ♤ ) = 3

{the queen, one ace and the nine are all spades}

● Total number of cards = 6

Probability of drawing a spade= 3/ 6

= 1/ 2

● Total number of "7" = 1

● Total number of cards = 6

Probability of drawing a 7

= 1/ 6

Now, what's asked is the difference in the probabilities of drawing a spade and a seven.

= 1/ 2 - 1/ 6

= 3/ 6 - 1/ 6

= 2/ 6

= 1/ 3

Hence, 1/ 3 of a greater chance of drawing a spade over a 7.

5 0

2 years ago