Problem 1: Toy Cars Weight

The task involves determining the weight difference between the heaviest toy car and each of the other toy cars, given the following information:

• Total weight of four toy cars: 3.85 lbs
• Weight of the heaviest car: 1 lb
• The remaining three toy cars each weigh the same amount.

To find the weight of each of the other three cars, we can use the formula for the total weight:

1. Let:
      - The weight of one of the other cars be denoted as $x$.
   

2. Write the equation for total weight:
      - The equation can be expressed as:
   $1 + 3x = 3.85$
      - Here, 1 is the weight of the heaviest car, and the term $3x$ accounts for the total weight of the other three cars.

3. Solve for x:
      - To solve for $x$, we can rearrange the equation:
   $3x = 3.85 - 1$
   $3x = 2.85$
   
      - Now dividing both sides by 3 gives:
   $x = \frac{2.85}{3}$
   $x = 0.95 ext{ lbs}$

4. Calculate the weight difference:
      - Now, we find the difference between the heaviest car and one of the other cars:
   $ext{Difference} = 1 - 0.95 = 0.05 ext{ lbs}$
   
   Thus, the heaviest car weighs 0.05 lbs more than one of the other toy cars.

Problem 2: Evaluate the Sum

This task involves evaluating the sum formula given by:

• $ext{Evaluate: } ext{Σ} (5k + 8)$
• With the index running from k = 1 to k = n (where n is unspecified in the transcript). Assuming the sum is for a specific upper limit, we will evaluate it for the given limit.

1. Evaluate for k=1:
      - When $k = 1$:
   $5(1) + 8 = 5 + 8 = 13$

2. Assuming an upper limit of n = 9:
      - This means we will keep adding for k = 1 to 9:
   $ext{Σ}<em>{k=1}^{9} (5k + 8) = ext{Σ}</em>{k=1}^{9}(5k) + ext{Σ}<em>{k=1}^{9}(8)$    - The first summation $ext{Σ}</em>{k=1}^{9} 5k$ can be computed as:
   $5 ext{Σ}<em>{k=1}^{9} k$    - The sum of the first n natural numbers is given by the formula:    $ext{Σ}</em>{k=1}^{n} k = \frac{n(n + 1)}{2}$
      - Thus, for n = 9:
   $ext{Σ}<em>{k=1}^{9} k = \frac{9(9 + 1)}{2} = \frac{9 imes 10}{2} = 45$    - Therefore, $ext{Σ}</em>{k=1}^{9} 5k = 5 imes 45 = 225$

3. Evaluate the constant summation:
      - Now, for the second summation:
   $ext{Σ}_{k=1}^{9} 8 = 8 imes 9 = 72$

4. Combine the results:
      - Finally, we can combine both parts:
   $ext{Σ}_{k=1}^{9} (5k + 8) = 225 + 72 = 297$

Therefore, the final sum evaluated is 297.