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qaws [65]
3 years ago
10

Suppose that farmers grew and unexpectedly large number of tomatoes this year .how would this increase in production affect the

price of tomatoes?
Mathematics
2 answers:
zloy xaker [14]3 years ago
3 0

Answer:

The price would go down as people would be more willing to buy cheaper potatoes and the producers would want to sell them for a lower price to get rid of them easy and fast.

Mariana [72]3 years ago
3 0

Answer:

The price of tomatoes would decrease

Step-by-step explanation: I aswell am doing Plato learning l

Ignore the rest

Ms. Thornton's class visited five freshwater lakes to learn more about the crocodiles and alligators living in them. The class counted the number of species in each lake as shown in the table.

Suppose that farmers grew an unexpectedly large number of tomatoes this year. How would this increase in production affect the price of tomatoes?

A.

The price of tomatoes would increase.

B.

OC. The

The price of tomatoes would decrease.

price of tomatoes would remain stable

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Alexeev081 [22]
Use Law of Cooling:
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T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
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Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
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The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
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