dalllc

2022-09-01

Find the sum of the two numbers. Give the numerator and denominator of your answer as an improper fraction in simple form $\frac{17}{42}+3\frac{5}{6}$

Find the difference between the two numbers. Give the numerator and denominator of your answer in simplest form.

$1\frac{1}{12}-\frac{2}{3}$

Determine the product of the fraction. Report the numerator and the denominator of your answer as a proper fraction in its simplest form. The only measured value is in numerator of the second fraction.

$\frac{5}{18}\cdot \frac{13}{19}$

Simplify the quotient. Report the numerator and the denominator of your answer as a fraction in its simplest form.

$\frac{5}{2}\xf7\frac{11}{3}$

Find the difference between the two numbers. Give the numerator and denominator of your answer in simplest form.

$1\frac{1}{12}-\frac{2}{3}$

Determine the product of the fraction. Report the numerator and the denominator of your answer as a proper fraction in its simplest form. The only measured value is in numerator of the second fraction.

$\frac{5}{18}\cdot \frac{13}{19}$

Simplify the quotient. Report the numerator and the denominator of your answer as a fraction in its simplest form.

$\frac{5}{2}\xf7\frac{11}{3}$

Ufumanaxi

Beginner2022-09-02Added 5 answers

a) $\frac{17}{42}+3\frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{17}{42}+\frac{23}{6}$

LCM of 6 and 42 is

$\frac{17+161}{42}\phantom{\rule{0ex}{0ex}}=\frac{178}{42}\phantom{\rule{0ex}{0ex}}=\frac{89}{21}$

numerator=89

denominator=21

b) $1\frac{1}{12}-\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{13}{12}=\frac{2}{3}$

LCM of 3 and 12 is

$\frac{13-8}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{12}$

numerator=5

denominator=12

c) $\frac{5}{18}\times \frac{13}{19}\phantom{\rule{0ex}{0ex}}=\frac{5\times 13}{18\times 19}\phantom{\rule{0ex}{0ex}}=\frac{65}{342}$

numerator=65

denominator=342

d) $\frac{5}{2}\xf7\frac{11}{3}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\times \frac{3}{11}\phantom{\rule{0ex}{0ex}}=\frac{15}{22}$

numerator=15

denominator=12

LCM of 6 and 42 is

$\frac{17+161}{42}\phantom{\rule{0ex}{0ex}}=\frac{178}{42}\phantom{\rule{0ex}{0ex}}=\frac{89}{21}$

numerator=89

denominator=21

b) $1\frac{1}{12}-\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{13}{12}=\frac{2}{3}$

LCM of 3 and 12 is

$\frac{13-8}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{12}$

numerator=5

denominator=12

c) $\frac{5}{18}\times \frac{13}{19}\phantom{\rule{0ex}{0ex}}=\frac{5\times 13}{18\times 19}\phantom{\rule{0ex}{0ex}}=\frac{65}{342}$

numerator=65

denominator=342

d) $\frac{5}{2}\xf7\frac{11}{3}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\times \frac{3}{11}\phantom{\rule{0ex}{0ex}}=\frac{15}{22}$

numerator=15

denominator=12

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