Complete the square to solve LaTeX: 0=-x^2+4x-5 Also written as 0= -x^2+4x-5

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

Which equation could be used to solve this problem? A number n is 8 less than the sum of 20 and 12. Find the number.

kifflom [539]

(20 + 12) - 8, since the question states n is EIGHT less than the sum of 20 and 12.

3 0

3 years ago

Identify the range of the function

NeX [460]

Answer:

what function

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Which inequality represents all values of X for which the product below is defined

balandron [24]

The domain for which the function is defined is given by:

D. $x \geq 4$ .

<h3>What is the domain of a function?</h3>

The domain of a function is the set that contains all possible input values for the function.

A square root is only defined for non-negative values, hence:

$x - 4 \geq 0 \rightarrow x \geq 4$ .

$x + 1 \geq 0 \rightarrow x \geq -1$ .

The intersection of these two domains is $x \geq 4$ , hence option D is correct.

More can be learned about the domain of a function at brainly.com/question/10891721

#SPJ1

6 0

2 years ago

For a club gathering you were having, you bought a total of 50 burgers, and spent $90. You paid $2 per turkey burger, and $1.50

Sedaia [141]

Answer:

No. You would have to cut the number of veggie burgers in more than half.

Explanation:

1. Model the situation with a system of equations

a) Name the variables:

number of turkey burgers: t
number of veggie burgers: v

b) Number of burgers:

50 = t + v

c) Cost of the 50 burgers:

$90 = 2t + 1.50v

2. Solve that system of equations:

a) System

50 = t + v

90 = 2t + 1.50v

b) Mutliply the first equation by 2 and subtract the second equation

100 = 2t + 2v
90 = 2t + 1.50v

10 = 0.5v

v = 20 ⇒ t = 50 - 20 = 30

c) How much would you spend if the next year you buy the double of 20 turkey burgers (40) and the half of 30 veggie burgers (15)

$2(40) + $1.50(15) = $80 + $22.50 = $102.50

Then, you if you double the number of turkey burgers, and cut the number burgers in half, you would spend more than $90 ($102.50).

6 0

4 years ago