Ask question

Ask question

All categories

2 years ago

12

Consider the expansion of (x + 2y)^10. Determine the coefficients for the terms with the powers of x and y shown.

2 answers:

julsineya [31]2 years ago

7 0

Answer with Step-by-step explanation:

We are given that an expression

$(x+2y)^{10}$

Binomial expansion o f $(x+y)^n$ is given by

= $nC_0 x^n+nC_1x^{n-1}y^1+nC_2x^{n-2}y^2+nC_3x^{n-3}y^3+... nC_n y^n$

a. We have to find the coefficient of $x^2y^8$

$(x+2y)^{10}=10C_0x^{10}+10C_1x^9(2y)+10C_2x^8(2y)^2+10C_3x^7(2y)^3+...+10C_{10}(2y)^{10}$

Term is $10C_8x^(2y)^8$

Coefficient = $10C_8\cdot 2^8$

b.Term in which $x^4y^6$ is $10C_6x^4(2y)^6$

Coefficient = $10C_4\cdot 2^6$

c.We have to find the coefficient of $x^5y^5$

Term is $10C_5x^5(2y)^5$

Coefficient = $10C_5\cdot 2^5$

kifflom [539]2 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Evaluate the quotient of 1.05 x 103 and 1 x 103: 1.05 x 109 1.05 x 106 1.05 x 100 1.05 x 10-6

kakasveta [241]

Answer:

HARD A F Sorry

8 0

2 years ago

Read 2 more answers

Shrina is selling cookie dough for her soccer team. She sold 2 tubs of

Inga [223]

Answer: 22

Step-by-step explanation:

She sold two tubs of Oatmeal Raisins and it's 6 dollars a tub so we can do 6*2 or 6+6 (doesn't matter). We get $12. Then, she also sells 2 tubs of Peanut Butter, and since it's $5 a tub, then we do 5*2 or 5+5 to get 10. We add 12 and 10 (12+10) and get 22.

I'm not sure if this is right because you added $22, $24, $20, and $11 and I'm not sure what the purposes of those are.

6 0

2 years ago

Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

5 0

2 years ago

Negative 9 plus 2 over 3 plus open parentheses 1 plus 1 close parenthesis squared

Alik [6]

Answer:

⁻ $\frac{19}{3}$

Step-by-step explanation:

⁻ $9 + \frac{2}{3} (1+1)^{2}$

⁻ $9 + \frac{2}{3}$ × $2^{2}$

⁻ $9 + \frac{2}{3}$ × $4$

⁻ $9$ $+$ $\frac{2 x 4}{3}$

⁻ $9 + \frac{8}{3}$

⁻ $\frac{19}{3}$

8 0

3 years ago

Read 2 more answers

Which equation has a constant of proportionality equal to 9 y=81/3x y=9x y=3x y==1/9x

aev [14]

Answer:

B) y = 9 x has Proportionality Constant = 9

Step-by-step explanation:

Two quantities P and Q are said to be PROPORTIONAL if and only if:

P ∝ Q ⇔ P = k Q ⇔ $k = \frac{P}{Q}$

Here, k = PROPORTIONALITY CONSTANT

Given : k = 9

Now, let us consider the given expressions in which x ∝y

y = 81/3 x

Here, $\frac{y}{x} = \frac{81}{3} = 27 \ne 9$

So, here Proportionality Constant ≠ 9

y = 9 x

Here, $\frac{y}{x} = 9$

So, here Proportionality Constant = 9

y = 3 x

Here, $\frac{y}{x} = 3 \ne 9$

So, here Proportionality Constant ≠ 9

y = 1/9 x

Here, $\frac{y}{x} = \frac{1}{9} \ne 9$

So, here Proportionality Constant ≠ 9

Hence, only y = 9 x has Proportionality Constant = 9

8 0

3 years ago