I see you've uploaded an image of a page from your textbook. It contains problems from Practice 6, focusing on evaluating expressions in scientific notation. Let's break down how to solve these problems.

Here’s a summary of the given values:

• $a = 1.3 \times 10^6$
• $b = 4.9 \times 10^{-2}$
• $c = 8.32 \times 10^4$
• $d = 7.6 \times 10^{-3}$
• $e = 5.32 \times 10^5$

For the first few problems, here's how we solve:

1. Problem a: $5a$

   Multiply $5 \times a = 5 \times 1.3 \times 10^6$ $5 \times 1.3 = 6.5, \quad \text{so the answer is } 6.5 \times 10^6.$

2. Problem b: $7b$

   Multiply $7 \times b = 7 \times 4.9 \times 10^{-2}$ $7 \times 4.9 = 34.3, \quad \text{so the answer is } 3.43 \times 10^1.$

3. Problem c: $b + d$

   To add numbers in scientific notation, convert them to the same exponent first. We have: $b = 4.9 \times 10^{-2}, \quad d = 7.6 \times 10^{-3}$ Convert $b$ to have the same exponent as $d$: $b = 49 \times 10^{-3}$ Now add ( 49 \times 10^{-3} + 7.6 \times 10^{-3} = 56.6 \times 10^{-3} = 5.66 \times 10^{-2}. ]

4. Problem d: $2e - 4c$

   Multiply the constants: $2e = 2 \times 5.32 \times 10^5 = 10.64 \times 10^5$ $4c = 4 \times 8.32 \times 10^4 = 33.28 \times 10^4 = 3.328 \times 10^5$ Now subtract: $10.64 \times 10^5 - 3.328 \times 10^5 = 7.312 \times 10^5.$

This is a brief outline of how to approach each problem. I can help solve the remaining ones too. Would you like me to continue or clarify any step?

Here are 5 questions to expand on this:

1. How do you add numbers in scientific notation when they have different exponents?
2. What are the steps to multiply two numbers in scientific notation?
3. Can you explain how to subtract numbers in scientific notation efficiently?
4. How does moving the decimal place affect the exponent in scientific notation?
5. How do we express the final answer correctly with significant figures in scientific notation?

Tip:

When multiplying or dividing numbers in scientific notation, remember to handle the exponents separately from the coefficients, which simplifies the calculation.