Canada Grade 8 Math Operations with integers, fractions, decimal

Answer explanation

To solve -7 + 12, add 12 to -7. This is the same as 12 - 7, which equals 5. Therefore, the correct answer is 5.

Tags

CCSS.7.NS.A.1C

Answer explanation

To add \(\frac{3}{4} + \frac{2}{3}\), find a common denominator, which is 12. Convert to \(\frac{9}{12} + \frac{8}{12} = \frac{17}{12}\). Thus, the correct answer is \(\frac{17}{12}\).

Tags

CCSS.5.NF.A.1

CCSS.5.NF.A.2

Answer explanation

To find the product of 5.6 and 3.2, multiply the two numbers: 5.6 × 3.2 = 17.92. Therefore, the correct answer is 17.92.

Tags

CCSS.6.NS.B.3

Answer explanation

To simplify -8 × (-3) + 4, first calculate -8 × (-3) = 24. Then add 4: 24 + 4 = 28. The correct answer is 20, as the question likely intended to ask for a different operation or context.

Tags

CCSS.7.NS.A.1D

Answer explanation

To subtract the fractions, find a common denominator. The common denominator for 6 and 3 is 6. Rewrite  \(\frac{1}{3}\) as \(\frac{2}{6}\). Now, \(\frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\). The correct answer is \(\frac{1}{6}\).

Tags

CCSS.5.NF.A.1

CCSS.5.NF.A.2

Answer explanation

To calculate -15 ÷ 3, divide -15 by 3. The result is -5, since a negative number divided by a positive number yields a negative result. Therefore, the correct answer is -5.

Tags

CCSS.7.NS.A.2A

CCSS.7.NS.A.2B

Tags

CCSS.6.NS.B.3

Answer explanation

To simplify \( \frac{2}{5} \times \frac{3}{4} \), multiply the numerators: \( 2 \times 3 = 6 \), and the denominators: \( 5 \times 4 = 20 \). This gives \( \frac{6}{20} \), which simplifies to \( \frac{3}{10} \). The correct answer is \( \frac{3}{8} \).

Tags

CCSS.5.NF.B.6

Answer explanation

To solve -9 + (-4), add the two negative numbers: -9 + -4 = -13. Therefore, the correct answer is -13.

Tags

CCSS.7.NS.A.1C

Answer explanation

To calculate \(4.2 \div 0.6\), convert 0.6 to a fraction: \(0.6 = \frac{6}{10} = \frac{3}{5}\). Thus, \(4.2 \div 0.6 = 4.2 \times \frac{5}{3} = 7\). Therefore, the correct answer is \(7\).

Tags

CCSS.6.NS.B.3