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Inessa [10]
3 years ago
13

Paul and Cindy own different pizza shops. Paul makes 12 pizzas the first day and 6 pizzas each day after that. Cindy makes 16 pi

zzas the first day and 4 pizzas each day after that. Who will have made more pizzas after 9 days? How many pizzas will that person have made? A. Paul. He has made 80 pizzas after 9 days. B. Cindy. She has made 132 pizzas after 9 days. C. Paul. He has made 60 pizzas after 9 days. D. Cindy. She has made 48 pizzas after 9 days.
Mathematics
1 answer:
eduard3 years ago
7 0

Answer:

Step-by-step explanation:

Paul : 12 pizzas the first day and 6 pizzas each day after that...for 9 days...keep in mind, we have the first day accounted for....so we will need 12 for the first day, and then we figure it for 8 days

ur equation would be : y = 12 + 6x...with x being the number of days and y being the total pizzas.......x would be 8..so we sub in 8 for x and find the total (y)

y = 12 + 6(8).....y = 12 + 48....y = 60....so paul made 60 pizzas in 9 days

Cindy : 16 pizzas the first day and 4 pizzas each day after that

y = 16 + 4x....x = 8

y = 16 + 4(8).....y = 16 + 32....y = 48.....cindy made 48 pizzas in 9 days

Paul would have made more pizzas

paul has made 60 pizzas in 9 days

cindy has made 48 pizzas in 9 days

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A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is
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The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Work done (Wd) =?

<h3>How to determine the spring constant</h3>
  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>
  • Spring constant (K) = 5 pound/foot
  • Extention (e) = 0.7 feet
  • Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

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