Question
a national park covers a area of 1.42 x 10^6 acres. A nearby national forest covers an area of 1.2 x 10^6 acres in the of 1.2 x 10^6 acres. What is the total number of acres in the park and the forest combined?
a. 1.54 x 10^6
b. 2.62 x 10^6
c. 15.5 x 10^6
d. 2.62 x 10^12
a. 1.54 x 10^6
b. 2.62 x 10^6
c. 15.5 x 10^6
d. 2.62 x 10^12
Answers
Answer
The distance from Venus to the sun is about 67,000,000 miles. The distance from Earth to the sun is about 2.6 x 10^7 farther. What is the distance from Earth to the sun? Write your answer in scientific notation to 1 decimal place.
Answer
calculate 4.8 x 10^-3 - 9.4 x 10^-6. Express the answer in scientific notation rounding to 2 decimal places.
Answer
a strand of hair has an average mass of 2.17 x 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.
___ x ___ ounces
___ x ___ ounces
Answer
Light travels at an approximate rate of 1.9 x 10^5 mi/sec. There are about 3.2 x 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.
Answer
The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2 x 10^6 years? Write the answer in scientific notation, expressed to the exact decimal place.
Answered by
GPT-4o mini
To find the total area covered by the national park and the national forest, you simply add the two areas together:
- Area of the national park = \(1.42 \times 10^6\) acres
- Area of the national forest = \(1.2 \times 10^6\) acres
Now add these two values:
\[
1.42 \times 10^6 + 1.2 \times 10^6 = (1.42 + 1.2) \times 10^6 = 2.62 \times 10^6
\]
So, the total number of acres in the park and the forest combined is \(2.62 \times 10^6\).
The correct answer is **b. 2.62 x 10^6**.
- Area of the national park = \(1.42 \times 10^6\) acres
- Area of the national forest = \(1.2 \times 10^6\) acres
Now add these two values:
\[
1.42 \times 10^6 + 1.2 \times 10^6 = (1.42 + 1.2) \times 10^6 = 2.62 \times 10^6
\]
So, the total number of acres in the park and the forest combined is \(2.62 \times 10^6\).
The correct answer is **b. 2.62 x 10^6**.
Answered by
GPT-4o mini
To find the distance from Earth to the sun, we need to add the distance from Venus to the sun to the additional distance from Earth to the sun.
1. The distance from Venus to the sun is **67,000,000 miles**.
2. The additional distance from Earth to the sun is about **2.6 x 10^7 miles**.
Let's first convert **67,000,000 miles** into scientific notation:
\[
67,000,000 = 6.7 \times 10^7 \text{ miles}
\]
Now, we can add the two distances together:
\[
\text{Distance from Earth to the sun} = 6.7 \times 10^7 + 2.6 \times 10^7
\]
Combining the coefficients:
\[
(6.7 + 2.6) \times 10^7 = 9.3 \times 10^7
\]
Thus, the distance from Earth to the sun, in scientific notation to one decimal place, is:
\[
\boxed{9.3 \times 10^7} \text{ miles}
\]
1. The distance from Venus to the sun is **67,000,000 miles**.
2. The additional distance from Earth to the sun is about **2.6 x 10^7 miles**.
Let's first convert **67,000,000 miles** into scientific notation:
\[
67,000,000 = 6.7 \times 10^7 \text{ miles}
\]
Now, we can add the two distances together:
\[
\text{Distance from Earth to the sun} = 6.7 \times 10^7 + 2.6 \times 10^7
\]
Combining the coefficients:
\[
(6.7 + 2.6) \times 10^7 = 9.3 \times 10^7
\]
Thus, the distance from Earth to the sun, in scientific notation to one decimal place, is:
\[
\boxed{9.3 \times 10^7} \text{ miles}
\]
Answered by
GPT-4o mini
To subtract \( 4.8 \times 10^{-3} - 9.4 \times 10^{-6} \), we first need to express both numbers with the same exponent.
The first number has an exponent of \(-3\), and the second has an exponent of \(-6\). We can convert \( 4.8 \times 10^{-3} \) to have the same exponent as \( 9.4 \times 10^{-6} \):
\[
4.8 \times 10^{-3} = 4.8 \times 10^{-3} \times \frac{10^3}{10^3} = 4.8 \times 10^{0} \times 10^{-6} = 4.8 \times 10^{3} \times 10^{-6} = 4800 \times 10^{-6}
\]
Now we can perform the subtraction:
\[
4800 \times 10^{-6} - 9.4 \times 10^{-6} = (4800 - 9.4) \times 10^{-6} = 4790.6 \times 10^{-6}
\]
Next, we convert \( 4790.6 \) into scientific notation.
\[
4790.6 = 4.7906 \times 10^{3}
\]
So, we can express the result as:
\[
4790.6 \times 10^{-6} = 4.7906 \times 10^{3} \times 10^{-6} = 4.7906 \times 10^{-3}
\]
Now rounding \( 4.7906 \) to two decimal places:
\[
4.79 \times 10^{-3}
\]
Thus, the final answer is:
\[
\boxed{4.79 \times 10^{-3}}
\]
The first number has an exponent of \(-3\), and the second has an exponent of \(-6\). We can convert \( 4.8 \times 10^{-3} \) to have the same exponent as \( 9.4 \times 10^{-6} \):
\[
4.8 \times 10^{-3} = 4.8 \times 10^{-3} \times \frac{10^3}{10^3} = 4.8 \times 10^{0} \times 10^{-6} = 4.8 \times 10^{3} \times 10^{-6} = 4800 \times 10^{-6}
\]
Now we can perform the subtraction:
\[
4800 \times 10^{-6} - 9.4 \times 10^{-6} = (4800 - 9.4) \times 10^{-6} = 4790.6 \times 10^{-6}
\]
Next, we convert \( 4790.6 \) into scientific notation.
\[
4790.6 = 4.7906 \times 10^{3}
\]
So, we can express the result as:
\[
4790.6 \times 10^{-6} = 4.7906 \times 10^{3} \times 10^{-6} = 4.7906 \times 10^{-3}
\]
Now rounding \( 4.7906 \) to two decimal places:
\[
4.79 \times 10^{-3}
\]
Thus, the final answer is:
\[
\boxed{4.79 \times 10^{-3}}
\]
Answered by
GPT-4o mini
To find out how much heavier a grain of rice is than a strand of hair, we will subtract the mass of the strand of hair from the mass of the grain of rice.
1. Mass of a strand of hair: \( 2.17 \times 10^{-5} \) ounces
2. Mass of a grain of rice: \( 0.002 \) ounces
First, we need to convert \( 0.002 \) ounces into scientific notation:
\[
0.002 = 2.0 \times 10^{-3} \text{ ounces}
\]
Now we can set up the subtraction:
\[
2.0 \times 10^{-3} - 2.17 \times 10^{-5}
\]
To perform the subtraction, we need a common exponent. The larger exponent is \(-3\), so we can convert \( 2.17 \times 10^{-5} \) to have the same exponent:
\[
2.17 \times 10^{-5} = 0.0217 \times 10^{-3} \text{ ounces}
\]
Now we can perform the subtraction:
\[
2.0 \times 10^{-3} - 0.0217 \times 10^{-3} = (2.0 - 0.0217) \times 10^{-3} = 1.9783 \times 10^{-3} \text{ ounces}
\]
Next, we need to express \( 1.9783 \) in scientific notation:
\[
1.9783 \times 10^{-3} \text{ is already in scientific notation, but we can express it as } 1.98 \times 10^{-3} \text{ ounces (rounded to two decimal places).}
\]
Thus, the difference in mass, expressed in scientific notation, is:
\[
\boxed{1.98 \times 10^{-3} \text{ ounces}}
\]
1. Mass of a strand of hair: \( 2.17 \times 10^{-5} \) ounces
2. Mass of a grain of rice: \( 0.002 \) ounces
First, we need to convert \( 0.002 \) ounces into scientific notation:
\[
0.002 = 2.0 \times 10^{-3} \text{ ounces}
\]
Now we can set up the subtraction:
\[
2.0 \times 10^{-3} - 2.17 \times 10^{-5}
\]
To perform the subtraction, we need a common exponent. The larger exponent is \(-3\), so we can convert \( 2.17 \times 10^{-5} \) to have the same exponent:
\[
2.17 \times 10^{-5} = 0.0217 \times 10^{-3} \text{ ounces}
\]
Now we can perform the subtraction:
\[
2.0 \times 10^{-3} - 0.0217 \times 10^{-3} = (2.0 - 0.0217) \times 10^{-3} = 1.9783 \times 10^{-3} \text{ ounces}
\]
Next, we need to express \( 1.9783 \) in scientific notation:
\[
1.9783 \times 10^{-3} \text{ is already in scientific notation, but we can express it as } 1.98 \times 10^{-3} \text{ ounces (rounded to two decimal places).}
\]
Thus, the difference in mass, expressed in scientific notation, is:
\[
\boxed{1.98 \times 10^{-3} \text{ ounces}}
\]
Answered by
GPT-4o mini
To determine how many miles light travels in a year, you need to multiply the speed of light by the number of seconds in a year.
1. Speed of light: \( 1.9 \times 10^5 \) miles/second
2. Seconds in a year: \( 3.2 \times 10^7 \) seconds
Now, we can perform the multiplication:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
\[
\text{Distance} = (1.9 \times 10^5) \times (3.2 \times 10^7)
\]
To multiply these two numbers, we multiply the coefficients and add the exponents:
\[
1.9 \times 3.2 = 6.08
\]
\[
10^5 \times 10^7 = 10^{5 + 7} = 10^{12}
\]
Combining these results, we have:
\[
\text{Distance} = 6.08 \times 10^{12} \text{ miles}
\]
Thus, the approximate distance light travels in a year, expressed in scientific notation, is:
\[
\boxed{6.08 \times 10^{12}} \text{ miles}
\]
1. Speed of light: \( 1.9 \times 10^5 \) miles/second
2. Seconds in a year: \( 3.2 \times 10^7 \) seconds
Now, we can perform the multiplication:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
\[
\text{Distance} = (1.9 \times 10^5) \times (3.2 \times 10^7)
\]
To multiply these two numbers, we multiply the coefficients and add the exponents:
\[
1.9 \times 3.2 = 6.08
\]
\[
10^5 \times 10^7 = 10^{5 + 7} = 10^{12}
\]
Combining these results, we have:
\[
\text{Distance} = 6.08 \times 10^{12} \text{ miles}
\]
Thus, the approximate distance light travels in a year, expressed in scientific notation, is:
\[
\boxed{6.08 \times 10^{12}} \text{ miles}
\]
Answered by
GPT-4o mini
To calculate how many meters a tectonic plate would move in \(2 \times 10^6\) years at a rate of \(0.006\) meters per year, we can use the formula:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
Substituting in the values:
- Speed of the tectonic plate = \(0.006 \, \text{m/year}\)
- Time = \(2 \times 10^6 \, \text{years}\)
Now we calculate:
\[
\text{Distance} = 0.006 \, \text{m/year} \times 2 \times 10^6 \, \text{years}
\]
First, multiply \(0.006\) and \(2\):
\[
0.006 \times 2 = 0.012
\]
Now, we have:
\[
\text{Distance} = 0.012 \times 10^6 \, \text{m}
\]
Next, convert \(0.012\) into scientific notation:
\[
0.012 = 1.2 \times 10^{-2}
\]
So, substituting this into the equation gives us:
\[
\text{Distance} = 1.2 \times 10^{-2} \times 10^6 = 1.2 \times 10^{4} \, \text{m}
\]
Thus, the tectonic plate would move approximately:
\[
\boxed{1.2 \times 10^4} \, \text{meters}
\]
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
Substituting in the values:
- Speed of the tectonic plate = \(0.006 \, \text{m/year}\)
- Time = \(2 \times 10^6 \, \text{years}\)
Now we calculate:
\[
\text{Distance} = 0.006 \, \text{m/year} \times 2 \times 10^6 \, \text{years}
\]
First, multiply \(0.006\) and \(2\):
\[
0.006 \times 2 = 0.012
\]
Now, we have:
\[
\text{Distance} = 0.012 \times 10^6 \, \text{m}
\]
Next, convert \(0.012\) into scientific notation:
\[
0.012 = 1.2 \times 10^{-2}
\]
So, substituting this into the equation gives us:
\[
\text{Distance} = 1.2 \times 10^{-2} \times 10^6 = 1.2 \times 10^{4} \, \text{m}
\]
Thus, the tectonic plate would move approximately:
\[
\boxed{1.2 \times 10^4} \, \text{meters}
\]