Intermediate Algebra Problems with Scientific Notation
Real-World Examples - Sample 5
This collection of intermediate algebra problems focuses on scientific notation, featuring practical examples commonly encountered in chemistry, engineering, physics, and
related fields. Detailed, step-by-step solutions are provided at the bottom of this page to help you build confidence and proficiency in handling scientific notation in real-world contexts.
Review
A number \( x \) is written in scientific notation when it is written in the form:
\[
x = a \times 10^N
\]
with
\[ 1 \leq |a| \lt 10 \]
and \( N \) is an integer.
Problems
-
Convert from decimal notation to scientific notation.
- \( 234 \)
- \( 0.000052 \)
- \( -220{,}000 \)
- \( -0.00000000000000182 \)
-
Convert from scientific to decimal notation.
- \( 1.01 \times 10^{2} \)
- \( 8.2 \times 10^{-6} \)
- \( 3.456 \times 10^{10} \)
- \( -1.2 \times 10^{-8} \)
-
Perform the multiplication and write the result in scientific notation.
- \( (5 \times 10^{2})(7 \times 10^{5}) \)
- \( (1.2 \times 10^{-6})(3.5 \times 10^{14}) \)
- \( (-4.5 \times 10^{-7})(8.62 \times 10^{10}) \)
- \( (-7.2 \times 10^{-6})(0.00039) \)
-
Perform the division and write the result in scientific notation.
- \( \dfrac{4 \times 10^{13}}{5 \times 10^{-4}} \)
- \( \dfrac{0.000012}{2.4 \times 10^{-8}} \)
- \( \dfrac{-1.2 \times 10^{-6}}{1.5 \times 10^{-10}} \)
- \( \dfrac{0.00000065}{0.000013} \)
-
Evaluate and express in scientific notation.
- \( (1.3 \times 10^{-6})^2 \)
- \( (1.5 \times 10^{-8})^2(2.1 \times 10^5) \)
- \( \dfrac{(1.144 \times 10^2)(6 \times 10^7)}{2.2 \times 10^4} \)
- \( \dfrac{(0.00001)(2{,}500)}{(0.0003)(150{,}000)} \)
-
Evaluate in scientific notation.
- Diameter of an atom: \( 0.0000000000001 \) meter
- Diameter of Earth: \( 12{,}756{,}000 \) meters
- 1 billion: \( 1{,}000{,}000{,}000 \)
- 1 micron: \( 0.000001 \) meter
- 1 trillion: \( 1{,}000{,}000{,}000{,}000 \)
- \( \dfrac{\text{Diameter of Earth}}{\text{Diameter of atom}} \)
- Speed of light: \( 300{,}000{,}000 \) m/sec
- Distance traveled by light in one year
Answers to the Above Questions
-
- \( 234 = 2.34 \times 10^2 \)
- \( 0.000052 = 5.2 \times 10^{-5} \)
- \( -220{,}000 = -2.2 \times 10^5 \)
- \( -0.00000000000000182 = -1.82 \times 10^{-15} \)
-
- \( 1.01 \times 10^2 = 101 \)
- \( 8.2 \times 10^{-6} = 0.0000082 \)
- \( 3.456 \times 10^{10} = 34{,}560{,}000{,}000 \)
- \( -1.2 \times 10^{-8} = -0.000000012 \)
-
-
\( (5 \times 10^2)(7 \times 10^5) = (5 \times 7)(10^2 \times 10^5) = 35 \times 10^7 = 3.5 \times 10^8 \)
-
\( (1.2 \times 10^{-6})(3.5 \times 10^{14}) = (1.2 \times 3.5)(10^{-6} \times 10^{14}) = 4.2 \times 10^8 \)
-
\( (-4.5 \times 10^{-7})(8.62 \times 10^{10}) = (-4.5 \times 8.62)(10^{-7} \times 10^{10}) = -38.79 \times 10^3 = -3.879 \times 10^4 \)
-
\( (-7.2 \times 10^{-6})(0.00039) = (-7.2 \times 10^{-6})(3.9 \times 10^{-4}) = (-7.2 \times 3.9)(10^{-6} \times 10^{-4}) = -28.08 \times 10^{-10} = -2.808 \times 10^{-9} \)
-
-
\( \dfrac{4 \times 10^{13}}{5 \times 10^{-4}} = \dfrac{4}{5} \times \dfrac{10^{13}}{10^{-4}} = 0.8 \times 10^{17} = 8.0 \times 10^{16} \)
-
\( \dfrac{0.000012}{2.4 \times 10^{-8}} = \dfrac{0.000012}{2.4} \times \dfrac{1}{10^{-8}} = 0.000005 \times 10^8 = 5.0 \times 10^2 \)
-
\( \dfrac{-1.2 \times 10^{-6}}{1.5 \times 10^{-10}} = \dfrac{-1.2}{1.5} \times \dfrac{10^{-6}}{10^{-10}} = -0.8 \times 10^4 = -8.0 \times 10^3 \)
-
\( \dfrac{0.00000065}{0.000013} = 0.05 = 5.0 \times 10^{-2} \)
-
-
\( (1.3 \times 10^{-6})^2 = 1.3^2 \times (10^{-6})^2 = 1.69 \times 10^{-12} \)
-
\( (1.5 \times 10^{-8})^2(2.1 \times 10^5) = 1.5^2 \times (10^{-8})^2 \times (2.1 \times 10^5) = (2.25 \times 10^{-16})(2.1 \times 10^5) = 4.725 \times 10^{-11} \)
-
\[
\dfrac{(1.144 \times 10^2)(6 \times 10^7)}{2.2 \times 10^4} = \dfrac{6.864 \times 10^9}{2.2 \times 10^4} = \dfrac{6.864}{2.2} \times \dfrac{10^9}{10^4} = 3.12 \times 10^5
\]
-
\[
\dfrac{(0.00001)(2500)}{(0.0003)(150000)} = \dfrac{0.025}{45} = 0.000555 = 5.6 \times 10^{-4}
\]
-
-
The diameter of an atom: \( 0.0000000000001 \text{ meter} = 1.0 \times 10^{-13} \text{ meter} \)
-
Diameter of Earth: \( 12{,}756{,}000 \text{ meters} = 1.2756 \times 10^7 \text{ meters} \)
-
1 billion: \( 1{,}000{,}000{,}000 = 1.0 \times 10^9 \)
-
1 micron: \( 0.000001 \text{ meter} = 1.0 \times 10^{-6} \)
-
1 trillion: \( 1{,}000{,}000{,}000{,}000 = 1.0 \times 10^{12} \)
-
\( \dfrac{\text{Diameter of Earth}}{\text{Diameter of atom}} = \dfrac{1.2756 \times 10^7}{1.0 \times 10^{-13}} = 1.2756 \times 10^{20} \)
-
Speed of light: \( 300{,}000{,}000 \text{ m/s} = 3.0 \times 10^8 \text{ m/s} \)
-
Distance travelled by light in one year:
Time in one year: \( 365 \times 24 \times 60 \times 60 = 3.1536 \times 10^7 \text{ seconds} \)
Distance = Time × Speed = \( (3.1536 \times 10^7)(3.0 \times 10^8) = 9.4608 \times 10^{15} \text{ meters} = 9.4608 \times 10^{12} \text{ kilometers} \)
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