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Go Math! 5 Common Core, Grade: 5 Publisher: Houghton Mifflin Harcourt

Go math 5 common core, title : go math 5 common core, publisher : houghton mifflin harcourt, isbn : 547587813, isbn-13 : 9780547587813, use the table below to find videos, mobile apps, worksheets and lessons that supplement go math 5 common core., textbook resources.

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Eureka Math Grade 5 Module 3 Lesson 6 Answer Key

Engage ny eureka math 5th grade module 3 lesson 6 answer key, eureka math grade 5 module 3 lesson 6 problem set answer key.

Explanation : The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{3}\) . 1\(\frac{2}{3}\) and  \(\frac{3}{5}\) have lcm 15 so, the rectangle is divided into 15 parts by drawing 4 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts. The left over shaded parts represent the difference of the given fraction to its total parts .

Question 2. Jean-Luc jogged around the lake in 1\(\frac{1}{4}\) hour. William jogged the same distance in \(\frac{5}{6}\) hour. How much longer did Jean-Luc take than William in hours? Answer: Time taken to jog for Jean-Luc = 1\(\frac{1}{4}\) hour . Time taken to jog for William = \(\frac{5}{6}\) hour . Time taken by Jean-Luc than William = 1\(\frac{1}{4}\) – \(\frac{5}{6}\)  = \(\frac{5}{4}\) – \(\frac{5}{6}\)  = \(\frac{30}{24}\)  – \(\frac{20}{24}\) = \(\frac{10}{24}\) = \(\frac{5}{12}\) hour .

Question 3. Is it true that 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\)? Prove your answer. Answer: Yes , 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\) Explanation : 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{7}{5}\) – \(\frac{3}{4}\) = \(\frac{28}{20}\) – \(\frac{15}{20}\) = \(\frac{13}{20}\) . \(\frac{1}{4}\) + \(\frac{2}{5}\) = \(\frac{5}{20}\) + \(\frac{8}{20}\) = \(\frac{13}{20}\) Therefore, 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\) = \(\frac{13}{20}\)  .

Eureka Math Grade 5 Module 3 Lesson 6 Exit Ticket Answer Key

Explanation : The Rectangle is divided into 3 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{3}\) . 1\(\frac{1}{3}\) and  \(\frac{5}{6}\) have lcm 6 so, the rectangle is divided into 6 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts. The left over shaded parts represent the difference of the given fraction to its total parts .

Eureka Math Grade 5 Module 3 Lesson 6 Homework Answer Key

Question 2. Sam had 1\(\frac{1}{2}\) m of rope. He cut off \(\frac{5}{8}\) m and used it for a project. How much rope does Sam have left? Answer: Length of Rope with Sam = 1\(\frac{1}{2}\) m Length of rope used for project = \(\frac{5}{8}\) m Length of Rope left = 1\(\frac{1}{2}\) – \(\frac{5}{8}\) = \(\frac{3}{2}\) – \(\frac{5}{8}\) = \(\frac{12}{8}\) – \(\frac{5}{8}\) = \(\frac{7}{8}\) . Therefore, Length of rope left with sam = \(\frac{7}{8}\) m .

Question 3. Jackson had 1\(\frac{3}{8}\) kg of fertilizer. He used some to fertilize a flower bed, and he only had \(\frac{2}{3}\) kg left. How much fertilizer was used in the flower bed? Answer: Quantity of fertilizers with Jackson = 1\(\frac{3}{8}\) kg Quantity of fertilizers left = \(\frac{2}{3}\) kg . Quantity of Fertilizers used for flower bed  = 1\(\frac{3}{8}\) kg  – \(\frac{2}{3}\) kg = \(\frac{11}{8}\) – \(\frac{2}{3}\) = \(\frac{33}{24}\) – \(\frac{16}{24}\) = \(\frac{17}{24}\) . Therefore, Quantity of Fertilizers used for flower bed  =\(\frac{17}{24}\)

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