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Ede4ka [16]
2 years ago
7

a person was born on may 14, 40 B.C. and died on may 14, 30 A.D. how many years did this person live.

Mathematics
1 answer:
Natali5045456 [20]2 years ago
7 0
This can be a trick question.

You can treat this problem as a problem of subtraction of integers.

Let B.C. years be represented by negative integers, and let A.D. years be represented by positive integers. Then you subtract.

40 B.C. is -4030 A.D. is 30


Then you have 30 - (-40) = 30 + 40 = 70

So far it's straightforward, and it seems the person lived for 70 years.


Here comes the (possible) trick with this question.

There was no year 0.

The counting of the years started with year 1, then year, 2, etc.

When we account for years before the year 1 A.D., the closest year to 1 A.D. is 1 B.C., not the year 0.

Since there is no year 0, then you must subtract 1 year from our calculation above, and the person lived 69 years.
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18cm

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4 = w

The rectangle is going to have two sides that are the length of l, and two sides that are the the length of w. We're looking for the perimiter p.

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√-225 • √-36 . show work.
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±90

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On the other hand, ...

... √(-225) · √(-36) = √((-225)·(-36)) = √8100 = 90

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If you consider all the roots at each stage, the result is ±90. Since you're working with complex numbers here, it is reasonable to recognize every number has two square roots.

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
olasank [31]

Answer:  a_n=2(5)^{n-2}

<u>Step-by-step explanation:</u>

The explicit rule for a geometric sequence is: a_n=a_1(r)^{n-1}\quad \text{where}\ a_1\ \text{is the first term and r is the ratio}

The information provided is: a₁ = \dfrac{2}{5}  and  r = 5

a_n=\dfrac{2}{5}(5)^{n - 1}\\\\.\quad = 2(5)^{n-2}

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2 years ago
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