Operations with scientific notation (ACT). Game on.

Operations with scientific notation (ACT) is a grade 8 math skill aligned to Common Core standard 8.EE.A.4. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 15 operations with scientific notation (act) problems our math games drill.

CCSS 8.EE.A.415 questions in the bank
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Warm-upeasy

What is (2×103)(3×104)(2 \times 10^{3})(3 \times 10^{4}) in scientific notation?

Multiply coefficients (2⋅3=62 \cdot 3 = 6) and add exponents: 3+4=7.3 + 4 = 7\textsf{.} So (2×103)(3×104)=6×107.(2 \times 10^{3})(3 \times 10^{4}) = 6 \times 10^{7}\textsf{.}

Mid-gameeasy

What is 9×1083×102\dfrac{9 \times 10^{8}}{3 \times 10^{2}} in scientific notation?

Divide coefficients (9÷3=39 \div 3 = 3) and subtract exponents: 8−2=6.8 - 2 = 6\textsf{.} So 9×1083×102=3×106.\dfrac{9 \times 10^{8}}{3 \times 10^{2}} = 3 \times 10^{6}\textsf{.}

Mid-gameeasy

What is (3×102)(3×103)(3 \times 10^{2})(3 \times 10^{3}) in scientific notation?

Coefficients multiply to 9;9\textsf{;} exponents add: 2+3=5.2 + 3 = 5\textsf{.} So (3×102)(3×103)=9×105.(3 \times 10^{2})(3 \times 10^{3}) = 9 \times 10^{5}\textsf{.}

Mid-gameeasy

8×1068 \times 10^{6} is how many times as large as 4×103?4 \times 10^{3}\textsf{?}

8×1064×103=2×103=2,000.\dfrac{8 \times 10^{6}}{4 \times 10^{3}} = 2 \times 10^{3} = 2{,}000\textsf{.}

Mid-gameeasy

9×1059 \times 10^{5} is how many times as large as 3×102?3 \times 10^{2}\textsf{?}

9×1053×102=3×103=3,000.\dfrac{9 \times 10^{5}}{3 \times 10^{2}} = 3 \times 10^{3} = 3{,}000\textsf{.}

Mid-gamemedium

What is (5×10−3)(4×10−2)(5 \times 10^{-3})(4 \times 10^{-2}) in scientific notation?

Multiply coefficients and add exponents: 5⋅4=205 \cdot 4 = 20 and −3+(−2)=−5-3+(-2)=-5, giving 20×10−520 \times 10^{-5}. Rewrite 20=2×10120=2 \times 10^{1}, so 2×101×10−5=2×10−4.2 \times 10^{1} \times 10^{-5}=2 \times 10^{-4}\textsf{.}

Mid-gamemedium

6.3×10−42.1×10−9=\dfrac{6.3 \times 10^{-4}}{2.1 \times 10^{-9}}=

Divide coefficients and subtract exponents: 6.3÷2.1=36.3 \div 2.1 = 3 and −4−(−9)=5-4-(-9)=5. So the quotient is 3×105.3 \times 10^{5}\textsf{.}

Buzzer beatermedium

(4×103)(5×104)(4 \times 10^{3})(5 \times 10^{4}) equals

Multiply coefficients and add exponents: 4⋅5=204 \cdot 5 = 20 and 3+4=73+4=7, giving 20×10720 \times 10^{7}. Rewrite as 2×101×107=2×108.2 \times 10^{1} \times 10^{7}=2 \times 10^{8}\textsf{.}

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