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shutvik [7]
3 years ago
14

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
2 answers:
lara [203]3 years ago
6 0

Answer:

70 years

Step-by-step explanation:

35 years before Christ and then another 35 years after:  35 + 35 = 70 years

yawa3891 [41]3 years ago
5 0

Answer:

the answer is 69

Step-by-step explanation:

35bc - 35ad = 69 years, because there is no year 0.

35+35 = 70

70-1=69

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The solution to this system of equations lies between the x-values -2 and -1.5. At which x-value are the two equations approxima
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Hello,
\left \{ {{y= \dfrac{1}{x+2} } \atop {y=x^2+2}} \right. \\\\

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For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10
professor190 [17]

The true statements are:

  • Tyson's expression is not equivalent to the original expression
  • The equivalent expression is:\frac{1}{2}g- \frac 1{5}

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}

Collect like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10}  -\frac{1}{10}

Evaluate the like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}

Tyler's equivalent expression is given as:

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}

Collect like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}

Evaluate the like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

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