180 Days of Math for Sixth Grade Day 78 Answers Key

By accessing our 180 Days of Math for Sixth Grade Answers Key Day 78 regularly, students can get better problem-solving skills.

180 Days of Math for Sixth Grade Answers Key Day 78

Directions: Solve each problem.

Question 1.
Calculate 52 and 29 more.
Answer:
180 Days of Math for Sixth Grade Answers Key Day 78_1

Question 2.
372 × 10 = ___________
Answer: 372 × 10 = 3720

Question 3.
Divide 437 by 9.
Answer:
437 divided by 9 is 48.5

Question 4.
Circle the prime number(s).
63
71
81
91
Answer:
180 Days of Math for Sixth Grade Answers Key Day 78_4 is the prime number
So, the correct answer is option B.

Question 5.
Is \(\frac{10}{12}\) less than, greater than, or equal to \(\frac{5}{6}\)?
Answer:
The simplified form of \(\frac{10}{12}\) is \(\frac{5}{6}\)
So, the fraction \(\frac{10}{12}\) is equal to \(\frac{5}{6}\)

Question 6.
2 × 5 – 4 × 3 = ____________
Answer:
(2 × 5) – (4 × 3)
10 – 12
-2

Question 7.
180 Days of Math for Sixth Grade Day 78 Answers Key 1
Answer:
We have to convert the given mixed fraction into the improper fraction.
2 \(\frac{2}{3}\) = \(\frac{8}{3}\)
So, 180-Days-of-Math-for-Sixth-Grade-Day-78-Answers-Key-7

Question 8.
Find f.
34f = 204
f = _________
Answer:
Given,
34f = 204
f = 204/34
f = 6

Question 9.
Calculate the area of a square with 12 cm sides.
Answer:
Given,
s = 12 cm
We know that,
Area of a square = s × s
A = 12cm × 12 cm
A = 144 sq.cm
Thus the area of a square is 144 sq. cm.

Question 10.
How many lines of symmetry does a rectangle have?
Answer: A rectangle has 2 lines of symmetry.

Question 11.
Inside a bag of candy there are 16 lollipops, 12 pieces of chocolate, and 7 pieces of licorice, If you reach into the bag and grab a piece of candy, what is the probability that it will not be a lollipop?
Answer:
Given data,
Inside a bag of candy there are 16 lollipops, 12 pieces of chocolate, and 7 pieces of licorice.
Total = 16 + 12 + 7 = 35
The probability that it will not be a lollipop = (16 + 12)/35
= 28/35

Question 12.
A burger restaurant serves twice as many cheeseburgers as hamburgers. It also sells \(\frac{1}{3}\) as many double hamburgers as regular hamburgers. If the restaurant sells 120 cheeseburgers, how many double hamburgers does it sell?
Answer:
Given,
A burger restaurant serves twice as many cheeseburgers as hamburgers.
It also sells \(\frac{1}{3}\) as many double hamburgers as regular hamburgers
120 × \(\frac{1}{3}\) = 40 double hamburgers

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