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Firlakuza [10]
1 year ago
9

Use the binomial expansion and approximation to find √3​

Mathematics
1 answer:
belka [17]1 year ago
8 0

According to the use of binomial expansion, the approximate value of √3 is found by applying the infinite sum √3 = 1 + (1 /2) · 2 - (1 / 8) · 2² + (1 / 16) · 2³ - (5 / 128) · 2⁴ + (7 / 256) · 2⁵ - (21 / 1024) · 2⁶ + (33 / 2048) · 2⁷ - (429 / 32768) · 2⁸ +...

An acceptable result cannot be found manually for it requires a <em>high</em> number of elements, with the help of a solver we find that the <em>approximate</em> value of √3 is 1.732.  

<h3>How to approximate the value of a irrational number by binomial theorem</h3>

Binomial theorem offers a formula to find the <em>analytical</em> form of the power of a binomial of the form (a + b)ⁿ:

(a + b)^{n} = \sum \limits_{k = 0}^{n} \left(\begin{array}{c}n\\k\end{array}\right)\cdot a^{n-k} \cdot b^{k}     (1)

Where:

  • a, b - Constants of the binomial.
  • n - Grade of the power binomial.
  • k - Index of the k-th element of the power binomial.

If we know that a = 1, b = 2 and n = 1 / 2, then an approximate expression for the square root is:

√3 = 1 + (1 /2) · 2 - (1 / 8) · 2² + (1 / 16) · 2³ - (5 / 128) · 2⁴ + (7 / 256) · 2⁵ - (21 / 1024) · 2⁶ + (33 / 2048) · 2⁷ - (429 / 32768) · 2⁸ +...

To learn more on binomial expansions: brainly.com/question/12249986

#SPJ1

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What does X equal in 4x+1=5x-4
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Step-by-step explanation:

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3 years ago
I can't figure out how to do (i + j) x (i x j)for vector calc
Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}

Then, solving the determinant

\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}

In our case,

\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}

Where we used the formula for AxB to calculate ixj.

Finally,

\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}

Thus, (i+j)x(ixj)=i-j

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1 year ago
Explain: What is the solution to this system of equations? <br><br> 2x+y=20 <br> 6x-5y=12
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Answer:

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Step-by-step explanation:

solving by substitution method

2x +y=20--------------1

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from equation 1, solve for y

2x+y=20

y= 20-2x------equation 3

adding value of y in equation 2

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adding value of x in equation 3

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so solution set (x,y) = (7,6)

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3 years ago
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I hope this helps you

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