Solved on Sep 07, 2023

Convert measurements to scientific notation: a. $1.2 \times 10^9 \mathrm{dL}$ b. $6.7 \times 10^{-11} \mathrm{mm}$ c. $6.7 \times 10^{10} \mathrm{g}$

STEP 1

Assumptions1. We are asked to write the given numbers in exponential notation. . Exponential notation is a way of writing a number that is too large or too small to be conveniently written in decimal form. It is written as a product of a number between1 and10 and a power of10.

STEP 2

Let's start with the first part (a). The number is $1200000000 \mathrm{d}$ .
We can write this number in exponential notation by identifying the number of places the decimal point needs to move to be between the first and second non-zero digits.

STEP 3

Move the decimal point9 places to the left to get $1.2$ . The number of places moved is the exponent of10.
So, $1200000000 \mathrm{d}$ can be written as $1.2 \times10^9 \mathrm{d}$ .

STEP 4

Next, let's move on to part (b). The number is $0.000000000067 \mathrm{mm}$ .
Again, we will move the decimal point, but this time to the right until it is between the first and second non-zero digits.

STEP 5

Move the decimal point11 places to the right to get $.7$ . The number of places moved is the negative of the exponent of10.
So, $0.000000000067 \mathrm{mm}$ can be written as $.7 \times10^{-11} \mathrm{mm}$ .

STEP 6

Finally, let's solve part (c). The number is $67000000000 \mathrm{g}$ .
As before, we will move the decimal point to the left until it is between the first and second non-zero digits.

STEP 7

Move the decimal point10 places to the left to get $6.7$ . The number of places moved is the exponent of10.
So, $67000000000 \mathrm{g}$ can be written as $6.7 \times10^{10} \mathrm{g}$ .
The solutions to the problem area. $1200000000 \mathrm{d} =1.2 \times10^9 \mathrm{d}$ b. $0.000000000067 \mathrm{mm} =6.7 \times10^{-11} \mathrm{mm}$ c. $67000000000 \mathrm{g} =6.7 \times10^{10} \mathrm{g}$ .

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Convert measurements to scientific notation: a. $1.2 \times 10^9 \mathrm{dL}$ b. $6.7 \times 10^{-11} \mathrm{mm}$ c. $6.7 \times 10^{10} \mathrm{g}$

STEP 1

Assumptions1. We are asked to write the given numbers in exponential notation. . Exponential notation is a way of writing a number that is too large or too small to be conveniently written in decimal form. It is written as a product of a number between1 and10 and a power of10.

STEP 2

Let's start with the first part (a). The number is $1200000000 \mathrm{d}$ .
We can write this number in exponential notation by identifying the number of places the decimal point needs to move to be between the first and second non-zero digits.

STEP 3

Move the decimal point9 places to the left to get $1.2$ . The number of places moved is the exponent of10.
So, $1200000000 \mathrm{d}$ can be written as $1.2 \times10^9 \mathrm{d}$ .

STEP 4

Next, let's move on to part (b). The number is $0.000000000067 \mathrm{mm}$ .
Again, we will move the decimal point, but this time to the right until it is between the first and second non-zero digits.

STEP 5

Move the decimal point11 places to the right to get $.7$ . The number of places moved is the negative of the exponent of10.
So, $0.000000000067 \mathrm{mm}$ can be written as $.7 \times10^{-11} \mathrm{mm}$ .

STEP 6

Finally, let's solve part (c). The number is $67000000000 \mathrm{g}$ .
As before, we will move the decimal point to the left until it is between the first and second non-zero digits.

STEP 7

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Convert measurements to scientific notation: a. 1.2×109dL1.2 \times 10^9 \mathrm{dL}1.2×109dL b. 6.7×10−11mm6.7 \times 10^{-11} \mathrm{mm}6.7×10−11mm c. 6.7×1010g6.7 \times 10^{10} \mathrm{g}6.7×1010g

STEP 1

Assumptions1. We are asked to write the given numbers in exponential notation. . Exponential notation is a way of writing a number that is too large or too small to be conveniently written in decimal form. It is written as a product of a number between1 and10 and a power of10.

STEP 2

Let's start with the first part (a). The number is 1200000000d1200000000 \mathrm{d}1200000000d.We can write this number in exponential notation by identifying the number of places the decimal point needs to move to be between the first and second non-zero digits.

STEP 3

Move the decimal point9 places to the left to get 1.21.21.2. The number of places moved is the exponent of10.So, 1200000000d1200000000 \mathrm{d}1200000000d can be written as 1.2×109d1.2 \times10^9 \mathrm{d}1.2×109d.

STEP 4

Next, let's move on to part (b). The number is 0.000000000067mm0.000000000067 \mathrm{mm}0.000000000067mm.Again, we will move the decimal point, but this time to the right until it is between the first and second non-zero digits.

STEP 5

Move the decimal point11 places to the right to get .7.7.7. The number of places moved is the negative of the exponent of10.So, 0.000000000067mm0.000000000067 \mathrm{mm}0.000000000067mm can be written as .7×10−11mm.7 \times10^{-11} \mathrm{mm}.7×10−11mm.

STEP 6

Finally, let's solve part (c). The number is 67000000000g67000000000 \mathrm{g}67000000000g.As before, we will move the decimal point to the left until it is between the first and second non-zero digits.

STEP 7

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Convert measurements to scientific notation: a. $1.2 \times 10^9 \mathrm{dL}$ b. $6.7 \times 10^{-11} \mathrm{mm}$ c. $6.7 \times 10^{10} \mathrm{g}$

Let's start with the first part (a). The number is $1200000000 \mathrm{d}$ .
We can write this number in exponential notation by identifying the number of places the decimal point needs to move to be between the first and second non-zero digits.

Move the decimal point9 places to the left to get $1.2$ . The number of places moved is the exponent of10.
So, $1200000000 \mathrm{d}$ can be written as $1.2 \times10^9 \mathrm{d}$ .

Next, let's move on to part (b). The number is $0.000000000067 \mathrm{mm}$ .
Again, we will move the decimal point, but this time to the right until it is between the first and second non-zero digits.

Move the decimal point11 places to the right to get $.7$ . The number of places moved is the negative of the exponent of10.
So, $0.000000000067 \mathrm{mm}$ can be written as $.7 \times10^{-11} \mathrm{mm}$ .

Finally, let's solve part (c). The number is $67000000000 \mathrm{g}$ .
As before, we will move the decimal point to the left until it is between the first and second non-zero digits.