12.

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

3 years ago

Solve the equation 3x-4=8-x

solmaris [256]

Answer:

x=3

Step-by-step explanation:

3x-4=8-x(Group like terms)

3x+x=8+4(Add both sides)

4x=12(After adding you will proceed to divide both sides by 4)

x=3(x is 3 because 4 can divide 12 3 times that's why we have x as equal to 3)

5 0

3 years ago

How many 'words' can be made from the name ESTABROK with no restrictions

Bumek [7]

The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

<h3>How to determine the number of ways</h3>

Given the word:

ESTABROK

Then n = 8

p = 6

The formula for permutation without restrictions

P = n! ( n - p + 1)!

P = 8! ( 8 - 6 + 1) !

P = 8! (8 - 7)!

P = 8! (1)!

P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1

P = 40, 320 ways

Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

Learn more about permutation here:

brainly.com/question/4658834

#SPJ1

6 0

2 years ago

How many milliliters are in 5 liters?

blsea [12.9K]

Answer:

There are 5000 mililitres in 5 litres

Therefore option A.

Hope u got the answer............

5 0

3 years ago

5(x-3)=x+13 what’s the value of x

Lera25 [3.4K]

Answer:

7

Step-by-step explanation:

5(x-3)=x+13

5x-15=x+13

-x +15

4x=28

divide 28/4

answer is 7

3 0

3 years ago

Read 2 more answers