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3 years ago

An ancient Greek was born on April 1st, 35 B.C. and died on April 1st, 35 A.D. How many years did he live?

Mathematics

1 answer:

masya89 [10]3 years ago

3 0

Answer:

69 years

Step-by-step explanation:

Data provided in the question

Born date of an Ancient Greek = April 1st 35 BC

Diet date of an Ancient Greek = Aril 1st 35 AD

Based on the above information

We can say that

35 + 35 = 70

We deduct 1 as there is no zero

So, it would be

= 70 - 1 year

= 69 years

Hence, An ancient greek lives 69 years and the same is to be considered

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Sonbull [250]

Answer: Angles A and C are vertical angles.

Step-by-step explanation:

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3 years ago

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aleksklad [387]

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3 years ago

How can you know if statement is true? <br> 1/2+3/8=4/10=2/5

mixas84 [53]

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3 years ago

According to 2017 Belgium Labor data information: Adult Population in 2017= 4500, Number of employed in 2017= 2800, and Number o

maks197457 [2]

The labor-force participation rate of Belgium in 2017 was 77.78%

Given:

Adult population = 4500

Number of employed in 2017 = <em>2800</em>

Number of Unemployed in 2017 = 700

Labor force participation rate = (employed + unemployed) / total population × 100

= (2800 + 700) / 4,500 × 100

= 3,500 / 4500 × 100

= 0.777777777777777 × 100

= 77.77777777777777%

Approximately, 77.78%

Given:

Adult Population in 2016 = 3500

Number of employed in 2016 = 1800

Number of Unemployed in 2016= 600

Unemployment rate = unemployed population/ total labor force × 100

= 600 / 3,500 × 100

= 0.171428571428571 × 100

= 17.14285714285714%

Approximately,

17.14 %

Therefore, the unemployment rate of Japan in 2016 was 17.14%

Learn more about unemployment rate:

brainly.com/question/13280244

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2 years ago

If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?

Keith_Richards [23]

$\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}$

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

$\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}$

so, the domain is all real numbers EXCEPT that one.

4 0

3 years ago