Let the length of the third side of the triangle be $x$ centimeters. Then $5 \frac{1}{10} + 7 \frac{1}{5} + x = 20.$ Converting all the fractions to twentieths, we get
\[5 \frac{2}{20} + 7 \frac{4}{20} + x = 20.\]We can rewrite this as
\[\frac{104}{20} + x = 20.\]Then $x = 20 - \frac{104}{20} = \frac{20 \cdot 20 - 104}{20} = \frac{296}{20} = \boxed{14.8}$ centimeters.
A ant walks all the way around a triangle drawn in the dirt. Two of the dimensions of the triangle are 5 1/10 centimeters and 7 1/5 centimeters. The total distance traveled by the ant is 20 centimeters. What is the length of the third side of the triangle?
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Let's start by labeling the sides of the triangle. Let $AB$ be the side with length $5 \frac{1}{10}$ centimeters and $AC$ be the side with length $7 \frac{1}{5}$ centimeters. We want to find the length of the third side, so let's call it $BC$, with length $x$ centimeters.
The total distance traveled by the ant is the sum of the lengths of all three sides of the triangle, which we know is $20$ centimeters. So we can write the equation:
\[AB + AC + BC = 20.\]
Substituting in the lengths of $AB$ and $AC$, we have:
\[5 \frac{1}{10} + 7 \frac{1}{5} + x = 20.\]
Next, let's convert the mixed fractions to improper fractions. We get:
\[\frac{51}{10} + \frac{36}{5} + x = 20.\]
To solve for $x$, we first add the two fractions:
\[\frac{51}{10} + \frac{36}{5} = \frac{51 \cdot 2}{10 \cdot 2} + \frac{36}{5} = \frac{102}{20} + \frac{36}{5}.\]
Now, we write $20$ as $\frac{20}{1}$ to have the same denominator:
\[\frac{102}{20} + \frac{36}{5} = \frac{102}{20} + \frac{36}{5} \cdot \frac{4}{4} = \frac{102}{20} + \frac{144}{20}.\]
Adding the fractions, we get:
\[\frac{102}{20} + \frac{144}{20} = \frac{102 + 144}{20} = \frac{246}{20}.\]
So, our equation becomes:
\[\frac{246}{20} + x = 20.\]
To solve for $x$, we isolate $x$ by subtracting $\frac{246}{20}$ from both sides:
\[x = 20 - \frac{246}{20}.\]
To simplify the right side, we write $20$ as $\frac{20}{1}$ to have the same denominator:
\[x = \frac{20}{1} - \frac{246}{20} = \frac{20}{1} \cdot \frac{20}{20} - \frac{246}{20} = \frac{400}{20} - \frac{246}{20} = \frac{400 - 246}{20} = \frac{154}{20}.\]
Simplifying the fraction, we get:
\[x = \frac{154}{20} = 7 \frac{14}{20}.\]
Finally, we can reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor (which is $2$):
\[x = 7 \frac{14}{20} = 7 \frac{14 ÷ 2}{20 ÷ 2} = 7 \frac{7}{10}.\]
So, the length of the third side $BC$ is $\boxed{7 \frac{7}{10}}$ centimeters.
The total distance traveled by the ant is the sum of the lengths of all three sides of the triangle, which we know is $20$ centimeters. So we can write the equation:
\[AB + AC + BC = 20.\]
Substituting in the lengths of $AB$ and $AC$, we have:
\[5 \frac{1}{10} + 7 \frac{1}{5} + x = 20.\]
Next, let's convert the mixed fractions to improper fractions. We get:
\[\frac{51}{10} + \frac{36}{5} + x = 20.\]
To solve for $x$, we first add the two fractions:
\[\frac{51}{10} + \frac{36}{5} = \frac{51 \cdot 2}{10 \cdot 2} + \frac{36}{5} = \frac{102}{20} + \frac{36}{5}.\]
Now, we write $20$ as $\frac{20}{1}$ to have the same denominator:
\[\frac{102}{20} + \frac{36}{5} = \frac{102}{20} + \frac{36}{5} \cdot \frac{4}{4} = \frac{102}{20} + \frac{144}{20}.\]
Adding the fractions, we get:
\[\frac{102}{20} + \frac{144}{20} = \frac{102 + 144}{20} = \frac{246}{20}.\]
So, our equation becomes:
\[\frac{246}{20} + x = 20.\]
To solve for $x$, we isolate $x$ by subtracting $\frac{246}{20}$ from both sides:
\[x = 20 - \frac{246}{20}.\]
To simplify the right side, we write $20$ as $\frac{20}{1}$ to have the same denominator:
\[x = \frac{20}{1} - \frac{246}{20} = \frac{20}{1} \cdot \frac{20}{20} - \frac{246}{20} = \frac{400}{20} - \frac{246}{20} = \frac{400 - 246}{20} = \frac{154}{20}.\]
Simplifying the fraction, we get:
\[x = \frac{154}{20} = 7 \frac{14}{20}.\]
Finally, we can reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor (which is $2$):
\[x = 7 \frac{14}{20} = 7 \frac{14 ÷ 2}{20 ÷ 2} = 7 \frac{7}{10}.\]
So, the length of the third side $BC$ is $\boxed{7 \frac{7}{10}}$ centimeters.
9 answers