Subtract 4.66×10−5−7.34×10−7

Question

(1 point)
Responses

−2.68×102
negative 2.68 times 10 squared

4.5866×10−5
4.5866 times 10 superscript negative 5 baseline

−4.5866×10−5
negative 4.5866 times 10 superscript negative 5 baseline

2.68×102

Answer 1

To subtract \( 4.66 \times 10^{-5} - 7.34 \times 10^{-7} \), we need to express both numbers with the same exponent.

Let's convert \( 4.66 \times 10^{-5} \) to a form with \( 10^{-7} \):

\[ 4.66 \times 10^{-5} = 4.66 \times 10^{-5} \times \frac{100}{100} = 4.66 \times 10^{-5} = 466 \times 10^{-7} \]

Now we can perform the subtraction:

\[ 466 \times 10^{-7} - 7.34 \times 10^{-7} = (466 - 7.34) \times 10^{-7} = 458.66 \times 10^{-7} \]

To express this in scientific notation, we convert \( 458.66 \) back to a number that is between 1 and 10:

\[ 458.66 \times 10^{-7} = 4.5866 \times 10^{-5} \]

So, the result of the subtraction \( 4.66 \times 10^{-5} - 7.34 \times 10^{-7} \) is:

\[ \boxed{4.5866 \times 10^{-5}} \]

Answer 2

The U.S. national debt was $2.9617×1013
in 2021 and $3.98×1011
in 1971. Find how much the debt increased in the 50-year period. (1 point)
Responses

$3.0015×1013
3.0015 dollars times 10 superscript 13 baseline

$1.0183×1024
1.0183 dollars times 10 superscript 24 baseline

$2.9219×1011
2.9219 dollars times 10 superscript 11 baseline

$2.9219×1013

Answer 3

To find how much the U.S. national debt increased from 1971 to 2021, we will perform the subtraction:

\[ \text{Debt in 2021} - \text{Debt in 1971} = 2.9617 \times 10^{13} - 3.98 \times 10^{11} \]

First, we need to express both numbers with the same exponent. Let's convert \( 3.98 \times 10^{11} \) to a form with \( 10^{13} \):

\[ 3.98 \times 10^{11} = 0.0398 \times 10^{13} \]

Now we can perform the subtraction:

\[ 2.9617 \times 10^{13} - 0.0398 \times 10^{13} = (2.9617 - 0.0398) \times 10^{13} = 2.9219 \times 10^{13} \]

So, the increase in the national debt over the 50-year period is:

\[ \boxed{2.9219 \times 10^{13}} \]

Answer 4

The size of a hydrogen atom is about 1.2×10−10
m across. The size of an oxygen atom is 0.000000000152 m across. What is the difference between the sizes of these atoms in scientific notation?(1 point)
Responses

3.2×10−9
m
3.2 times 10 superscript negative 9 baseline m

3.2×10−11
m
3.2 times 10 superscript negative 11 baseline m

3.2 ×10−10
m
3.2 times 10 superscript negative 10 baseline m

2.72×10−10
m

Answer 5

To find the difference between the sizes of a hydrogen atom and an oxygen atom, first express both sizes in scientific notation:

1. Hydrogen atom: \( 1.2 \times 10^{-10} \) m
2. Oxygen atom: \( 0.000000000152 \) m, which can be expressed in scientific notation as: \[ 0.000000000152 = 1.52 \times 10^{-10} \text{ m} \]

Now, we can calculate the difference:

\[ \text{Size difference} = \text{Size of Hydrogen} - \text{Size of Oxygen} \] \[ = (1.2 \times 10^{-10}) - (1.52 \times 10^{-10}) \]

To perform the subtraction, we can do:

\[ 1.2 - 1.52 = -0.32 \]

Now, we express \( -0.32 \times 10^{-10} \) in scientific notation.

To convert \( -0.32 \) into scientific notation, we write it as:

\[ -0.32 = -3.2 \times 10^{-1} \]

So, combined with \( 10^{-10} \):

\[ -0.32 \times 10^{-10} = -3.2 \times 10^{-1} \times 10^{-10} = -3.2 \times 10^{-11} \text{ m} \]

However, as we are looking for the absolute difference in size, we take the absolute value, hence:

\[ \text{Difference} = 3.2 \times 10^{-11} \text{ m} \]

Thus, the answer is:

\[ \boxed{3.2 \times 10^{-11} \text{ m}} \]