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aniya is taking a road trip. The distance she plans to travel can be written as 3×10^2

kilometers, 3×10^5
meters, 3×10^7
centimeters, or 3×10^8
millimeters. Which distance uses the most appropriate units?(1 point)

5 answers

To determine which distance uses the most appropriate units, we should convert all the distances to the same unit and see which unit makes the most sense based on the context (a road trip).

  1. Kilometers:
    \(3 \times 10^2 \text{ kilometers} = 300 \text{ kilometers}\)

  2. Meters:
    \(3 \times 10^5 \text{ meters} = 300,000 \text{ meters}\)
    (This is equivalent to \(300\) kilometers, since \(1 \text{ kilometer} = 1000 \text{ meters}\)).

  3. Centimeters:
    \(3 \times 10^7 \text{ centimeters} = 30,000,000 \text{ centimeters}\)
    (This is equivalent to \(300\) kilometers, since \(1 \text{ kilometer} = 100,000 \text{ centimeters}\)).

  4. Millimeters:
    \(3 \times 10^8 \text{ millimeters} = 300,000,000 \text{ millimeters}\)
    (This is equivalent to \(300\) kilometers, since \(1 \text{ kilometer} = 1,000,000 \text{ millimeters}\)).

All distances are equivalent to 300 kilometers. However, the most appropriate unit for a road trip is kilometers, as it is a standard unit for measuring longer distances and is more practical than meters, centimeters, or millimeters.

Thus, the most appropriate unit is 3 × 10² kilometers.

The distance from Venus to the sun is about 67,000,000 miles. The distance from Earth to the sun is about 2.6×10^7
farther. What is the distance from Earth to the sun? Write your answer in scientific notation to 1 decimal place.(1 point)

Calculate 4.8×10^−3 − 9.4×10^−6
. Express the answer in scientific notation rounding to 2 decimal places.(1 point

A strand of hair has an average mass of 2.17×10−5
ounces. A grain of rice has an average mass of 0.002 ounces. How much heavier is a grain of rice than a strand of hair? Express your answer in scientific notation.(1 point)

Let's solve each of the problems one by one.

1. Distance from Earth to the Sun

The distance from Venus to the sun is \( 67,000,000 \) miles, and the distance from Earth to the sun is \( 2.6 \times 10^7 \) miles farther than that.

Step 1: Convert \( 67,000,000 \) miles to scientific notation
\( 67,000,000 \) can be expressed as \( 6.7 \times 10^7 \) miles.

Step 2: Calculate the distance from Earth to the sun
Earth to the sun distance = Venus to sun distance + Earth to Venus distance
Earth to the sun distance = \( 6.7 \times 10^7 , \text{miles} + 2.6 \times 10^7 , \text{miles} \)

To add these: \[ 6.7 \times 10^7 + 2.6 \times 10^7 = (6.7 + 2.6) \times 10^7 = 9.3 \times 10^7 \]

Final Answer:
The distance from Earth to the sun is approximately \( \mathbf{9.3 \times 10^7} \) miles.

2. Calculate \( 4.8 \times 10^{-3} - 9.4 \times 10^{-6} \)

First, we need to express \( 4.8 \times 10^{-3} \) and \( 9.4 \times 10^{-6} \) with the same power of ten:

Convert to same power of ten: \[ 4.8 \times 10^{-3} = 4.8 \times 10^{-3} , (no change) \ 9.4 \times 10^{-6} = 0.0000094 = 0.0000094 \times 10^3 = 9.4 \times 10^{-6} \times 10^3 = 0.0000094 \times 10^{-3} \]

Now we have: \( 4.8 \times 10^{-3} - 0.0000094 \times 10^{-3} = (4.8 - 0.0000094) \times 10^{-3} \).

Calculate:
\[ 4.8 - 0.0000094 \approx 4.7999906 \ \text{So the result is } \approx 4.7999906 \times 10^{-3} \]

Rounding to 2 decimal places:
\[ \approx 4.8 \times 10^{-3} \]

Final Answer:
The answer is \( \mathbf{4.8 \times 10^{-3}} \).

3. How much heavier is a grain of rice than a strand of hair?

Given:

  • Mass of a strand of hair = \( 2.17 \times 10^{-5} \) ounces
  • Mass of a grain of rice = \( 0.002 \) ounces = \( 2 \times 10^{-3} \) ounces

Step 1: Calculate the difference in mass
\[ \text{Difference} = \text{Mass of grain of rice} - \text{Mass of strand of hair} = 2 \times 10^{-3} - 2.17 \times 10^{-5} \]

Convert \( 2 \times 10^{-3} \) to the same power of ten:
\[ 2 \times 10^{-3} = 2.00 \times 10^{-3} \] \[ 2.00 \times 10^{-3} = 2.00 \times 10^{-3} = 20 \times 10^{-5} \] Now, we can subtract: \[ = (20 \times 10^{-5} - 2.17 \times 10^{-5}) = (20 - 2.17) \times 10^{-5} = 17.83 \times 10^{-5} \]

Step 2: Convert to scientific notation:
\[ 17.83 \times 10^{-5} = 1.783 \times 10^{-4} \]

Final Answer:
A grain of rice is \( \mathbf{1.78 \times 10^{-4}} \) ounces heavier than a strand of hair (rounded to two decimal places).

Summary of Answers:

  1. Distance from Earth to the Sun: \( \mathbf{9.3 \times 10^7} \) miles
  2. Calculation result: \( \mathbf{4.8 \times 10^{-3}} \)
  3. Difference in mass: \( \mathbf{1.78 \times 10^{-4}} \) ounces heavier
Light travels at an approximate rate of 1.9×10^5
mi./sec. There are about 3.2×10^7
seconds in a year. About how many miles does light travel in a year? Write the answer in scientific notation. Express the answer to the exact decimal place.(1 point)
The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×10^6
years? Write the answer in scientific notation, expressed to the exact decimal place.(1 point)

An asteroid is heading toward Mars at the rate of 7.7×10^4
mi./hr. If the asteroid is 3.311×10^8
miles away from Mars, how many hours will it be before it hits Mars? Write the answer in scientific notation, expressed to the exact decimal place.(1 point)
A field of 3.9 mi^2
is being subdivided into 1.3×10^3
family gardens. How many square miles will there be in each garden? Write the answer in scientific notation.(1 point)

Let's solve each of the problems one by one.

1. Distance Light Travels in a Year

To find out how far light travels in a year, we need to calculate:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

Given:

  • Speed of light = \( 1.9 \times 10^5 \) mi/sec
  • Time in a year = \( 3.2 \times 10^7 \) seconds

Step 1: Calculate the distance

\[ \text{Distance} = (1.9 \times 10^5 , \text{mi/sec}) \times (3.2 \times 10^7 , \text{sec}) \]

Using the property of exponents:

\[ \text{Distance} = 1.9 \times 3.2 \times 10^{5 + 7} \]

Step 2: Calculate the multiplication of the coefficients

\[ 1.9 \times 3.2 = 6.08 \]

Step 3: Combine the results

\[ \text{Distance} = 6.08 \times 10^{12} , \text{miles} \]

Final Answer: The distance light travels in a year is approximately \( \mathbf{6.08 \times 10^{12}} \) miles.

2. Distance a Tectonic Plate Moves in \( 2 \times 10^6 \) Years

Given:

  • Speed of tectonic plate = \( 0.006 \) m/year
  • Time = \( 2 \times 10^6 \) years

Step 1: Calculate the total distance

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

\[ \text{Distance} = 0.006 , \text{m/year} \times (2 \times 10^6 , \text{years}) \]

Step 2: Perform the calculation

Convert \( 0.006 \) to scientific notation: \[ 0.006 = 6 \times 10^{-3} \]

Now we compute:

\[ \text{Distance} = (6 \times 10^{-3}) \times (2 \times 10^6) = 12 \times 10^{3} = 1.2 \times 10^{4} \text{ m} \]

Final Answer: The tectonic plate would move \( \mathbf{1.2 \times 10^{4}} \) meters in \( 2 \times 10^6 \) years.

3. Time for the Asteroid to Hit Mars

Given:

  • Speed of the asteroid = \( 7.7 \times 10^{4} \) mi/hr
  • Distance to Mars = \( 3.311 \times 10^{8} \) miles

Step 1: Calculate the time until it hits Mars

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] \[ \text{Time} = \frac{3.311 \times 10^{8}}{7.7 \times 10^{4}} \]

Step 2: Perform the calculation

Using the property of exponents, \[ \text{Time} = \frac{3.311}{7.7} \times 10^{8 - 4} \]

Step 3: Calculate \( \frac{3.311}{7.7} \)

\[ \frac{3.311}{7.7} \approx 0.430 \]

So now: \[ \text{Time} \approx 0.430 \times 10^{4} = 4.30 \times 10^{3} , \text{hours} \]

Final Answer: The time until the asteroid hits Mars is approximately \( \mathbf{4.30 \times 10^{3}} \) hours.

4. Area Each Family Garden Will Cover

Given:

  • Total field area = \( 3.9 , \text{mi}^2 \)
  • Number of family gardens = \( 1.3 \times 10^{3} \)

Step 1: Calculate the area of each garden

\[ \text{Area per garden} = \frac{\text{Total Area}}{\text{Number of Gardens}} \] \[ \text{Area per garden} = \frac{3.9}{1.3 \times 10^{3}} \]

Step 2: Convert \( 3.9 \) to scientific notation

\[ 3.9 = 3.9 \times 10^0 \]

Now compute:

\[ \text{Area per garden} = \frac{3.9 \times 10^{0}}{1.3 \times 10^{3}} = \frac{3.9}{1.3} \times 10^{0 - 3} \]

Computing \( \frac{3.9}{1.3} \)

\[ \frac{3.9}{1.3} \approx 3.0 \]

So we have:

\[ \text{Area per garden} \approx 3.0 \times 10^{-3} , \text{mi}^2 \]

Final Answer: The area in each garden is approximately \( \mathbf{3.0 \times 10^{-3}} \) mi².

Summary of Answers:

  1. Light travel in a year: \( \mathbf{6.08 \times 10^{12}} \) miles
  2. Tectonic plate movement: \( \mathbf{1.2 \times 10^{4}} \) m
  3. Time for asteroid to hit Mars: \( \mathbf{4.30 \times 10^{3}} \) hours
  4. Area per family garden: \( \mathbf{3.0 \times 10^{-3}} \) mi²