Question

Challenge Add and subtract as indicated. Round off when necessary. a. $(4.32\times {10}^3\mathrm{cm})-(1.6\times {10}^6\mathrm{mm})$ b. $(2.12\times {10}^7\mathrm{mm})+(1.8\times {10}^3\mathrm{cm})$

Short answer

a. $-1.168\times {10}^{6}mm$ b. $2.1218\times {10}^{7}mm$

Step-by-step solution

Convert cm to mm

To change cm to mm, we multiply by 10: $$4.32\times {10}^{3}cm\times \frac{10mm}{1cm}=4.32\times {10}^{4}mm$$ Step 2: Subtract using scientific notation

Subtract the numbers in scientific notation

Now that we have the same units, we can subtract the numbers: $$(4.32\times {10}^{4}mm)-(1.6\times {10}^{6}mm)$$ To do this subtraction, we need the exponents of the numbers to be the same. Rewrite 4.32 x 10^4 as 0.432 x 10^6: $$(0.432\times {10}^{6}mm)-(1.6\times {10}^{6}mm)$$ Step 3: Subtract the coefficients

Subtract coefficients and simplify

We can now perform the subtraction since the exponents are the same: $$(0.432-1.6)\times {10}^{6}mm=-1.168\times {10}^{6}mm$$ #Problem a answer: $-1.168\times {10}^{6}mm$ #Problem b:# Step 1: Convert units

Convert cm to mm

To change cm to mm, we multiply by 10: $$1.8\times {10}^{3}cm\times \frac{10mm}{1cm}=1.8\times {10}^{4}mm$$ Step 2: Add using scientific notation

Add the numbers in scientific notation

Now that we have the same units, we can add the numbers: $$(2.12\times {10}^{7}mm)+(1.8\times {10}^{4}mm)$$ To do this addition, we need the exponents of the numbers to be the same. Rewrite 1.8 x 10^4 as 0.0018 x 10^7: $$(2.12\times {10}^{7}mm)+(0.0018\times {10}^{7}mm)$$ Step 3: Add the coefficients

Add coefficients and simplify

We can now perform the addition since the exponents are the same: $$(2.12+0.0018)\times {10}^{7}mm=2.1218\times {10}^{7}mm$$ #Problem b answer: $2.1218\times {10}^{7}mm$

Key concepts

Unit Conversion

Unit conversion is the process of converting a given measurement into another unit without changing its actual value. It's crucial in comparing measurements that are expressed in different units, allowing for uniform calculations. For instance, in the given problem, you need to convert centimeters (cm) to millimeters (mm) to make the units consistent before performing addition or subtraction.

To convert from centimeters to millimeters, you multiply by 10. This is because there are 10 millimeters in a centimeter. So,

• To convert 4.32 $\times {10}^3\text{cm}$ to mm, the conversion would be $4.32\times {10}^3\text{cm}\times \frac{10\text{mm}}{1\text{cm}}=4.32\times {10}^4\text{mm}$.
• Similarly, $1.8\times {10}^3\text{cm}$ becomes $1.8\times {10}^4\text{mm}$ after conversion.

Converting units ensures that you're working in the same measurement system, thereby making calculations straightforward.

Addition and Subtraction in Scientific Notation

Scientific notation is a way of expressing very large or very small numbers in a compact form. Each number is written as a product of two factors: a decimal greater than or equal to 1 and less than 10, and an integer power of ten. For instance, $4.32\times {10}^3$ and $1.6\times {10}^6$ are in scientific notation.

Addition and subtraction in scientific notation require the exponents to be the same. This is because you can only add or subtract the coefficients (the numbers in front) when they are paired with the same power of ten. If the exponents aren't the same, you must adjust one of the numbers to match the exponent of the other. Here's how:

• To subtract $4.32\times {10}^4\text{mm}$ from $1.6\times {10}^6\text{mm}$, rewrite $4.32\times {10}^4$ to $0.432\times {10}^6$.
• Perform the subtraction: $(0.432-1.6)\times {10}^6=-1.168\times {10}^6$.
• For addition, like with $2.12\times {10}^7\text{mm}$ and $1.8\times {10}^4\text{mm}$, you rewrite $1.8\times {10}^4$ as $0.0018\times {10}^7$.
• Add them to get $(2.12+0.0018)\times {10}^7=2.1218\times {10}^7$.

Always ensure the exponents are aligned before performing these operations.

Rounding Numbers

Rounding numbers is a mathematical technique used to simplify numbers, making them easier to work with by reducing the number of digits. It involves increasing or maintaining the value of the trailing digit depending on how close it is to a specific value.

In scientific notation, rounding is particularly useful to keep the calculations manageable and results concise. Here's a simple guideline:

• If the digit following the last significant figure is 5 or greater, increase the last significant figure by one.
• If it is less than 5, simply drop the trailing digits.

For instance, in the result $-1.168\times {10}^6$, if the context asks, you might round this to $-1.17\times {10}^6$ when fewer significant figures are required. Rounding makes results more readable and often conforms to the level of precision required by the problem or context.