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pishuonlain [190]
4 years ago
7

What is the mathematical notation for absolute value?​

Mathematics
2 answers:
alukav5142 [94]4 years ago
4 0
Use vertical pipes around the value.

Example:

abs(-4) = |-4| = 4
olchik [2.2K]4 years ago
4 0
It’s these vertical lines ||.


For instance to write the absolute value of -4, you write |-4|
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Step-by-step explanation:

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Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1
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Answer:

<em />S_{1893} =5632.98<em />

<em />

Step-by-step explanation:

The correct form of the question is:

S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

S_n = \frac{a(1 - r^n)}{1 - r}

But first, we need to get the number of terms in the sequence using:

T_n = ar^{n-1}

Where

a = First\ Term

a = 1.001^{1892}

r = common\ ratio

r = \frac{1}{1.001}

T_n = Last\ Term

T_n = 1

So, we have:

T_n = ar^{n-1}

1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}

Apply law of indices:

1 = 1.001^{1892} * (1.001^{-1})^{n-1}

1 = 1.001^{1892} * (1.001)^{-n+1}

Apply law of indices:

1 = 1.001^{1892-n+1}

1 = 1.001^{1892+1-n}

1 = 1.001^{1893-n}

Represent 1 as 1.001^0

1.001^0 = 1.001^{1893-n}

They have the same base:

So, we have

0 = 1893-n

Solve for n

n = 1893

So, there are 1893 terms in the sequence given.

Solving further:

S_n = \frac{a(1 - r^n)}{1 - r}

Where

a = 1.001^{1892}

r = \frac{1}{1.001}

n = 1893

So, we have:

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }

Simplify the numerator

S_{1893} =\frac{1.001^{1892}  -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892}  -1.001^{1892-1893}}{\frac{0.001}{1.001} }

S_{1893} =\frac{1.001^{1892}  -1.001^{-1}}{\frac{0.001}{1.001} }

S_{1893} =(1.001^{1892}  -1.001^{-1})/({\frac{0.001}{1.001} })

S_{1893} =(1.001^{1892}  -1.001^{-1})*{\frac{1.001}{0.001}}

S_{1893} =\frac{(1.001^{1892}  -1.001^{-1}) * 1.001}{0.001}

Open Bracket

S_{1893} =\frac{1.001^{1892}* 1.001  -1.001^{-1}* 1.001 }{0.001}

S_{1893} =\frac{1.001^{1892+1}  -1.001^{-1+1}}{0.001}

S_{1893} =\frac{1.001^{1893}  -1.001^{0}}{0.001}

S_{1893} =\frac{1.001^{1893}  -1}{0.001}

S_{1893} =5632.97970294

Hence, the sum of the sequence is:

<em />S_{1893} =5632.98<em> ----- approximated</em>

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