Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 2 Lesson 2.2 Adding and Subtracting in Scientific Notation to finish your assignments.

## Math in Focus Grade 8 Course 3 A Chapter 2 Lesson 2.2 Answer Key Adding and Subtracting in Scientific Notation

### Math in Focus Grade 8 Chapter 2 Lesson 2.2 Guided Practice Answer Key

**Complete.**

Question 1.

The population of Washington, D.C., is about 5.9 • 10^{5}. South Dakota has a population of approximately 8 • 10^{5}.

a) Find the sum of the populations.

Sum of populations

= Population of Washington D.C. + Population of South Dakota

The sum of the populations is about .

Answer:

Sum of populations

= Population of Washington D.C. + Population of South Dakota

5.9 • 10^{5} + 8 • 10^{5}.

Substitute the value 10^{5}. from each term

Add within the parentheses

(5.9 + 8) • 10^{5}.

13.9 • 10^{5}.

write 13.9 in the scientific notation.

1.39 • 10^{1}.• 10^{5}.

Use the product of powers property.

1.39 • 10^{6}.

b) Find the difference in the populations.

Difference in the populations

= Population of South Dakota – Population of Washington, D.C.

The difference in the populations is about .

Answer:

Population of South Dakota – Population of Washington, D.C.

8 • 10^{5} – 5.9 • 10^{5}.

Substitute the value 10^{5}. from each term

Add within the parentheses

(8 – 5.9) • 10^{5}.

2.1 • 10^{5}.

write 13.9 in the scientific notation.

21 • 10^{1}.• 10^{5}.

Use the product of powers property.

21 • 10^{6}.

Question 2.

The approximate length of the smallest salamander is 1.7 • 10^{-2} meter. The smallest lizard is about 1.6 • 10^{-2} meter long.

a) What is the sum of the lengths of the salamander and the lizard?

Sum of the lengths of the salamander and the lizard

= Length of salamander + Length of lizard

= 1.7 • 10^{-2} + 1.6 • 10^{-2} Substitute.

= ( + ) • Factor from each term.

= • m within parentheses.

The sum of the lengths is about meter.

Answer:

Sum of the lengths of the salamander and the lizard

= Length of salamander + Length of lizard

= 1.7 • 10^{-2} + 1.6 • 10^{-2} Substitute.

= (1.7 + 1.6) • 10^{-2}

= (3.3)• 10^{-2}

The sum of the lengths is about (3.3)• 10^{-2} meter.

b) What is the difference ¡n the length of the reptiles?

The salamander is about meter longer than the lizard.

Answer:

Difference in length between the salamander and lizard.

= Length of salamander – Length of lizard

= 1.7 • 10^{-2} – 1.6 • 10^{-2} Substitute.

= (1.7 – 1.6) • 10^{-2}

= 0.1 • 10^{-2}

= 1 • 10^{-1} • 10^{-2}

= 1 • 10^{-3}

Question 3.

The approximate area of the continent of Australia is 9 • 10^{6} square kilometers.

The area of the continent of Antarctica is about 1,37 • 10^{7} square kilometers.

a) Find the approximate sum of the land areas of the two continents.

The sum of the land areas is about square kilometers.

Answer:

Approximate sum of the land areas of the two continents

Area of Australia + Area of Antarctica

9 • 10^{6} + 1.37 • 10^{7
}9 • 10^{6} + 1.37 • 10^{6} • 10^{1
}(9 + 13.7) • 10^{6
}22.7 • 10^{6
}2.27 • 10^{7} square kilometers

b) What is the difference in the areas of the two continents?

Difference in the land areas

= Area of Antarctica – Area of Australia

The land area of Antarctica is about square kilometers larger than the land area of Australia.

Answer:

The difference in the land areas

= Area of Antarctica – Area of Australia

9 • 10^{6} – 1.37 • 10^{7
}1.37 • 10^{6} – 9 • 10^{6
}(13.7 – 9) • 10^{6
}4.7 • 10^{6 }square kilometers

**Solve. Write your answers in scientific notation.**

Question 4.

A custom-made invitation using a 10-pt card stock is about 2.54 • 10^{-4} meter thick. It is placed inside a tissue paper insert that is approximately 6.0 • 10^{-6} meter thick.

a) How thick is the invitation when placed in the tissue paper insert?

Answer:

2.54 • 10^{-4}+ 6.0 • 10^{-6} meters thick

2.54 • 10^{-4}+ 0.06 • 10^{-4} meters thick

(2.54 + 0.06) • 10^{-4} meters thick

2.6 • 10^{-4} meters

b) How much thicker is the invitation than the tissue paper insert?

Answer:

Let x is a variable here representing the number of times thicker.

Thickness of the card stock is x times as thick as tissue.

2.54 • 10^{-4}= x • 6.0 • 10^{-6}

x = 2.54 • 10^{-4}/6.0 • 10^{-6}

x = 0.423 • 10^{2}

x = 4.23 • 10^{2}

**Complete.**

Question 5.

On average, Pluto orbits the Sun at a distance of approximately 4,802 gigameters. Uranus’s average distance from the Sun is about 2.992 • 10^{9} kilometers. Which planet is farther from the Sun? How much farther?

Distance of Pluto from the Sun:

Gm = • km Write in scientific notation.

Difference in distance of Pluto and Uranus from the Sun

= Distance of from the Sun – Distance of from the Sun

= • – • Substitute.

= ( – ) • Factor from each term.

= • km within parentheses.

So, is kilometers farther from the Sun.

Answer:

Given,

On average, Pluto orbits the Sun at a distance of approximately 4,802 gigameters.

Uranus’s average distance from the Sun is about 2.992 • 10^{9} kilometers.

1 gigameter = 10^{6} kilometers.

4,802 gigameters = 4,802 • 10^{6} kilometers.

Difference in distance of Pluto and Uranus from the Sun

= Distance of Pluto from the Sun – Distance of Uranus from the Sun

4,802 • 10^{6} kilometers – 2.992 • 10^{9} kilometers

= 4,802 • 10^{6} kilometers – 2,992 • 10^{6} kilometers

= (4802 – 2992) • 10^{6} kilometers

= 1810 • 10^{6}

So, Pluto is 1810 • 10^{6} kilometers farther from the Sun.

### Math in Focus Course 3A Practice 2.2 Answer Key

**Solve. Show your work. Round the coefficient to the nearest tenth.**

Question 1.

6.3 • 10^{-2} + 4.9 • 10^{-2}

Answer:

(6.3 + 4.9) • 10^{-2}

(11.2) • 10^{-2}

Now, Round the coefficient to the nearest tenth.

11 • 10^{-2}

Question 2.

7.2 • 10^{2} – 3.5 • 10^{2}

Answer:

7.2 • 10^{2} – 3.5 • 10^{2}

(7.2 – 3.5) • 10^{2}

3.7 • 10^{2}

Now, Round the coefficient to the nearest tenth.

4 • 10^{2}

Question 3.

3.8 • 10^{3} + 5.2 • 10^{4}

Answer:

3.8 • 10^{3} + 5.2 • 10^{4}

3.8 • 10^{3} + 5.2 • 10^{3 }• 10^{1}

(3.8 + 52) • 10^{3}

55.8 • 10^{3}

Now, Round the coefficient to the nearest tenth.

56 • 10^{3}

Question 4.

8.1 • 10^{5} – 2.8 • 10^{4}

Answer:

8.1 • 10^{5} – 2.8 • 10^{4}

81 • 10^{4} – 2.8 • 10^{4}

(81 – 2.8) • 10^{4}

78.2 • 10^{4}

Now, Round the coefficient to the nearest tenth.

78 • 10^{4}

**Use the table to answer questions 5 to 9.**

The table shows the amounts of energy, in Calories, contained in various foods.

Question 5.

Find the total energy in each food combination. Write your answer in scientific notation. Round coefficients to the nearest tenth.

a) Chicken breast and cabbage

Answer:

Chicken breast: 1.71 • 10^{4}

cabbage: 2.5 • 10^{4}

Total energy in each food combination is

1.71 • 10^{4 }+ 2.5 • 10^{4}

(1.7 + 2.5) • 10^{4}

4.2 • 10^{4}

4 • 10^{4}

b) Cabbage and raw potato

Answer:

cabbage: 2.5 • 10^{4}

Raw potato: 7.7 • 10

Total energy in each food combination is

2500 • 10 + 7.7 • 10

2507.7 • 10

2510 • 10

Question 6.

How many more Calories are in chicken breast than in salmon?

Answer:

Chicken breast: 1.71 • 10^{4}

Salmon: 1.67 • 10^{5}

1.67 • 10^{5 }– 1.71 • 10^{4}

(16.7 – 1.71) • 10^{4}

14.99 • 10^{4}

Question 7.

How many more Calories are in salmon than in cabbage?

Answer:

Salmon: 1.67 • 10^{5}

cabbage: 2.5 • 10^{4}

1.67 • 10^{5 }– 2.5 • 10^{4}

(16.7 – 2.5) • 10^{4}

14.2 • 10^{4}

**Solve. Show your work.**

Question 8.

A flight from Singapore to New York includes a stopover at Hawaii. The distance between Singapore and Hawaii is about 6.7 • 10^{3} miles. The distance between New York and Hawaii is about 4.9 • 10^{3} miles. Write each sum or difference in scientific notation.

a) Find the total distance from Singapore to New York.

Answer:

The distance between Singapore and Hawaii is about 6.7 • 10^{3} miles.

The distance between New York and Hawaii is about 4.9 • 10^{3} miles.

6.7 • 10^{3} miles + 4.9 • 10^{3} miles

(6.7 + 4.9) • 10^{3} miles

11.6 • 10^{3} miles

b) Find the difference in the length of the two flights.

Answer:

The distance between Singapore and Hawaii is about 6.7 • 10^{3} miles.

The distance between New York and Hawaii is about 4.9 • 10^{3} miles.

6.7 • 10^{3} miles – 4.9 • 10^{3} miles

(6.7 – 4.9) • 10^{3} miles

1.8 • 10^{3} miles

Question 9.

Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^{-6} meter. Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^{-5} meter. Write your answers in the appropriate unit in prefix form.

a) Find the sum of the diameters of the two types of fiber.

Answer:

Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^{-6} meter.

Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^{-5} meter.

1.45 • 10^{-5} meter + 1 • 10^{-6} meter.

14.5 • 10^{-6} meter + 1 • 10^{-6} meter.

(14.5 + 1) • 10^{-6} meter.

15.5 • 10^{-6} meter.

1.55 • 10^{-5} meter.

b) How much wider is the cashmere fiber than the angora fiber?

Answer:

Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^{-6} meter.

Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^{-5} meter.

1.45 • 10^{-5} meter – 1 • 10^{-6} meter.

14.5 • 10^{-6} meter – 1 • 10^{-6} meter.

(14.5 – 1) • 10^{-6} meter.

13.5 • 10^{-6} meter.

1.35 • 10^{-5} meter.

**The average distances of three planets from the Sun are shown in the diagram. Use this information for questions 10 to 13. Express your answers in kilometers.**

Question 10.

What is the closest Mercury comes to Earth when both are at an average distance from the Sun?

Answer:

Mercury: 5.83 • 10^{10
}Earth: 1.5 • 10^{8} km = 1.5 • 10^{8} • 10^{3} m = 1.5 • 10^{10}• 10^{1
}Difference between the distance of Earth from Sun to distance of Mercury from Sun

15 • 10^{10} – 5.83 • 10^{10
}(15 – 5.83) • 10^{10
}9.17 • 10^{10}

Question 11.

What is the closest Saturn comes to Mercury when both are at an average distance from the Sun?

Answer:

Mercury: 5.83 • 10^{10}

Saturn: 1.43 • 10^{12}

Difference between the distance of Saturn from Sun to distance of Mercury from Sun

143 • 10^{10} – 5.83 • 10^{10
}(143 – 5.83) • 10^{10
}137.17 • 10^{10}

Question 12.

What is the closest Saturn comes to Earth when both are at an average distance from the Sun?

Answer:

Earth: 1.5 • 10^{8} km = 1.5 • 10^{8} • 10^{3} m = 1.5 • 10^{11}

Saturn: 1.43 • 10^{12}

Difference between the distance of Saturn from Sun to distance of Mercury from Sun

14.3 • 10^{11} – 1.5 • 10^{11
}(14.3 – 1.5) • 10^{11
}12.8• 10^{11}

Question 13.

Is the distance you found in 12 greater or less than the average distance from Earth to the Sun? Explain.

Answer: Yes the distance of Saturn from the Sun to the distance of Mercury from the Sun is 12.8• 10^{11}

**Solve. Show your work.**

Question 14.

Factories A and B produce potato chips. They use the same basic ingredients: potatoes, oil, and salt. Last year, each factory used different amounts of these ingredients, as shown in the table.

a) Which factory used more potatoes last year? How many more potatoes did it use?

Answer: By seeing the above table we can say that Factory A has used more potatoes last year than Factor B.

Potato Factory A Used: 4.87 • 10^{6}

Potato Factory B Used: 3,309 • 10^{3}

Potato Factory A Used: 4870 • 10^{3}

Potatoes Used Last Year: 4870 • 10^{3}– 3,309 • 10^{3}

(4870 – 3309) • 10^{3}

1561 • 10^{3}

b) Which factory used more oil last year? How much more oil did it use than the other factory?

Answer:

By seeing the above table we can say that Factory B has used more oil last year than Factor A.

Oil Factory A Used: 356,000 = 356 • 10^{3}

Oil Factory B Used: 5.61 • 10^{5 }= 561 • 10^{3}

Oil Used Last Year: 561 • 10^{3}– 356• 10^{3}

(561- 356) • 10^{3}

205• 10^{3}

c) Find the total weight of the ingredients used by each factory. Write your answer in scientific notation.

Answer:

Factory A:

Potato Factory A Used: 4870 • 10^{3}

Oil Factory A Used: 356,000 = 356 • 10^{3}

Salt Factory A Used: 2.87 • 10^{5 }= 287 • 10^{3}

4870 • 10^{3 }+ 356 • 10^{3} + 287 • 10^{3}

(4870 + 356 + 287) • 10^{3}

5513 • 10^{3}

Factory B:

Potato Factory B Used: 3,309 • 10^{3}

Oil Factory B Used: 561 • 10^{3}

Salt Factory B Used: 193500 = 193.5 • 10^{3 }

3309 • 10^{3 }+ 561 • 10^{3} + 193.5 • 10^{3}

(3309+ 561 + 193.5) • 10^{3}

4063.5 • 10^{3}

Question 15.

**Math journal** The approximate population of the following countries in North America in 2011 are shown in the table. Explain how to use scientific notation to find the total population of the countries.

Answer:

Mexico: 110,000,000 = 11 • 10^{7
}Haiti: 9,700 000 = 97 • 10^{5
}Costa Rica: 4,600,000 = 46 • 10^{5
}United States: 310,000,000 = 31 • 10^{7
}11 • 10^{7} + 31 • 10^{7} + 97 • 10^{5} + 46 • 10^{5
}4 • 10^{7} + 143 • 10^{5}