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snow_lady [41]
2 years ago
14

1%7D%7B1%2Bx%7D%2B%5Cfrac%7B2%7D%7B1%2Bx%5E%7B2%7D%7D%2B%5Cfrac%7B2%5E%7B2%7D%7D%7B1%2Bx%5E%7B4%7D%7D%2B%5Cldots%20%5Ccdot%20%5Cfrac%7B2%5E%7B100%7D%7D%7B1%2Bx%5E%7B200%7D%7D%5Cend%7Bequation%7D" id="TexFormula1" title="\begin{equation}\text { Question: Find the value of } \frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{2^{2}}{1+x^{4}}+\ldots \cdot \frac{2^{100}}{1+x^{200}}\end{equation}" alt="\begin{equation}\text { Question: Find the value of } \frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{2^{2}}{1+x^{4}}+\ldots \cdot \frac{2^{100}}{1+x^{200}}\end{equation}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ankoles [38]2 years ago
7 0

\bold{Heya!}

Your answer to this is:

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

<h2>→ <u>EXPLANATION :-</u></h2>

<u />

<u />\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: (1)

\sf{Apply \: rule:} \: (a) = a

\sf{(1) = 1

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: ^. \: 1

\sf{\frac{1}{1 \: + \:1} = \frac{1}{2}

\sf{\frac{2^1^0^0}{1 \: + \: x^2^0^0} ^.^ \: 1 \: = \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

Hopefully This Helps ! ~

#LearnWithBrainly

\underline{Answer :}

<em>Jaceysan ~</em>

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UNO [17]

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Answer:

  A. 3×3

  B. [0, 1, 5]

  C. (rows, columns) = (# equations, # variables) for matrix A; vector x remains unchanged; vector b has a row for each equation.

Step-by-step explanation:

A. The matrix A has a row for each equation and a column for each variable. The entries in each column of a given row are the coefficients of the corresponding variable in the equation the row represents. If the variable is missing, its coefficient is zero.

This system of equations has 3 equations in 3 variables, so matrix A has dimensions ...

  A dimensions = (rows, columns) = (# equations, # variables) = (3, 3)

Matrix A is 3×3.

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B. The second row of A represents the second equation:

  0x_1+1x_2+5x_3=-1

The coefficients of the variables are 0, 1, 5. These are the entries in row 2 of matrix A.

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C. As stated in part A, the size of matrix A will match the number of equations and variables in the system. If the number of variables remains the same, the number of rows of A (and b) will reflect the number of equations. (The number of columns of A (and rows of x) will reflect the number of variables.)

6 0
3 years ago
Complete the table for y=1/2x-3
Readme [11.4K]

Tables are created by substitute a set of input values into the function to create outputs. The required table is as shown below

<em>x | y</em>

<em>0   -3 </em>

<em>1   -2.5</em>

<em>2   -2 </em>

<h3>Tables and values</h3>

Tables are created by substitute a set of input values into the function to create outputs

Using x = 0, 1 and 2 as the input values

Given the function

y = 1/2x - 3

If x = 0

y = 1/2(0) - 3

y = -3

If x = 1

y = 1/2(1) - 3

y = -2.5

If x = 2

y = 1/2(2) - 3

y = -2

Hence the required table is as shown below

x | y

0   -3

1   -2.5

2   -2

Learn more on tables and values here: brainly.com/question/12151322

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8 0
2 years ago
(r/s)(3) = ?<br><br> please help.
MissTica

Answer:

\frac{3r}{s}

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3 years ago
The Mongolian Giant Sunflower produces seeds that are over 1 inch long, grow up to 18 feet tall, and flowers that are 1 ½ feet w
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Answer:

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Step-by-step explanation:

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2 years ago
Rewrite the expression using properties of exponents:
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