180 Days of Math for Third Grade Day 116 Answers Key

By accessing our 180 Days of Math for Third Grade Answers Key Day 116 regularly, students can get better problem-solving skills.

180 Days of Math for Third Grade Answers Key Day 116

Directions: Solve each problem.

Question 1.
180 Days of Math for Third Grade Day 116 Answers Key 1
Answer:
180-Days-of-Math-for-Third-Grade-Day-116-Answers-Key-1
By performing the addition operation for the two given numbers we can find the sum of 17 and 43
17 + 43 = 60

Question 2.
2 × 10 = 180 Days of Math for Third Grade Day 116 Answers Key 2
Answer: 20
Explanation:
By performing a multiplication operation we can find the product of 2 and 10.
So, by multiplying 2 and 10 we get 20.

Question 3.
2 × 11 = 180 Days of Math for Third Grade Day 116 Answers Key 2
Answer:22
Explanation:
By performing a multiplication operation we can find the product of 2 and 11.
So, by multiplying 2 and 11 we get 22.

Question 4.
Write the numeral for eighty.
Answer: The numeral for eighty is 80.

Question 5.
Circle the smaller fraction.
\(\frac{3}{10}\)
\(\frac{6}{10}\)
Answer:
In this case, the denominators of both the fractions are the same.
So, we have to compare the numerators to find the smaller fraction.
3 is less than 6.
So, \(\frac{3}{10}\) is smaller than \(\frac{6}{10}\).

Question 6.
67 + 180 Days of Math for Third Grade Day 116 Answers Key 2 = 73
Answer:
Let the unknown value be x
67 + x = 73
x = 73 – 67
x = 6
So, 67 + 6 = 73

Question 7.
100 cm = __________ m
Answer:
Convert from centimeters to meters
1 meter = 100 cm
So, 100 cm = 1 m

Question 8.
Is a door taller or shorter than one toot?
Answer: Door is taller than one toot.

Question 9.
Does the arrow point to a face, vertex, or edge?
180 Days of Math for Third Grade Day 116 Answers Key 3
Answer: The arrow in the above figure points to a face.

Question 10.
How much fencing is needed to enclose a yard with the dimensions shown below?
180 Days of Math for Third Grade Day 116 Answers Key 4
Answer:
Given,
L = 20 ft
W = 15 ft
We know that,
Perimeter of a rectangle = 2 (L + W)
P = 2 (20 + 15)
P = 2(35)
P = 70 feet
Thus the fencing needed to enclose a yard with the dimensions shown is 70 feet.

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