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fomenos
4 years ago
13

John is building a rectangular puppy kennel up against his house using 31 feet of fencing. The side against the house does not n

eed a fence and the side parallel to his house needs to be 15 feet long. Write an equation that models the situation and solve for the length of one of the shorter sides that extend out from the house.
Mathematics
1 answer:
Korolek [52]4 years ago
4 0
John want to build a rectangular puppy kennel up against his house. The side against the house doesn't need a fence, so John still need to build 3 fences. He got 31 feet of fence. the fence parallel to the side against the house should be 15 feet long. So he still need to determine the side of the 2 other fences, which are equal because its a rectangular (the parallel sides in a rectangular are equal).
So the equation that models the situation is:
31 = 2s + 15, in which s represents the length of the one fence.

We need to solve for s. Let's subtract 15 from each side to have the variables on a side and the numbers on the other:
31 = 2s + 15
31 - 15 = 2s + 15 - 15
16 = 2s

Then divide both sides by 2 to have the variable s on a side and its value on the other:
16 = 2s
16/2 = (2s)/2
8 = s.

So the shorter sides of the rectangle needs to be 8 feet long each.

Let's check our answer:
2s + 15 = 16 + 15 = 31.
Our answer has been approved.

Hope this helps! :D
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\dfrac{\dbinom{13}8\dbinom{13}2}{\dbinom{52}{10}}\approx0.0000063455

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3 years ago
Solve −8x ≤ 2. <br> A. x ≥ −.25<br> B. x ≥ −.30<br> C. x ≤ −.25<br> D. x ≤ .25
4vir4ik [10]
The answer would be A because when you divide by a negative, it flips the sign.
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3 years ago
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Answer:

The number of loaves to be made daily to maximize Mr. Joaquim's average profit is 1400.

O número de pães a serem feitos diariamente para maximizar o lucro médio do Sr. Joaquim é 1400.

Step-by-step explanation:

The table gives the daily demand of bread and the average profit that corresponds to each demand.

The daily demand values recorded ranged from 1000 to 1800.

And the corresponding average profit for the demands for each daily demand provided showed a steady increase from $0.10 as at a daily demand of 1000, through $0.15, then peaking at $0.20 at a daily demand of 1400.

After this point, the average profit begins to decline again as we move further away from the peak obtained at a daily demand of 1400.

Hence, it is evident that the daily demand that corresponds to the maximum average profit is 1400.

A daily demand of 1400 is where the peak average profit of $0.20 is obtained.

Any other demand lower or higher than 1400 corresponds to a departure from the peak profit obtained at that point according to the table of daily demand and average profits provided in the question.

In portugese/Em português

A tabela fornece a demanda diária de pão e o lucro médio que corresponde a cada demanda.

Os valores diários de demanda registrados variaram de 1000 a 1800.

E o lucro médio correspondente às demandas de cada demanda diária fornecida mostrou um aumento constante de US $ 0,10, com uma demanda diária de 1000, passando por US $ 0,15, atingindo um pico de US $ 0,20 com uma demanda diária de 1400.

Após esse ponto, o lucro médio começa a declinar novamente à medida que nos afastamos do pico obtido com uma demanda diária de 1400.

Portanto, é evidente que a demanda diária que corresponde ao lucro médio máximo é de 1.400.

Uma demanda diária de 1400 é onde é obtido o lucro médio máximo de US $ 0,20.

Qualquer outra demanda menor ou maior que 1400 corresponde a um afastamento do pico de lucro obtido nesse momento, de acordo com a tabela de demanda diária e lucros médios fornecidos na pergunta.

Hope this Helps!!!

Espero que isto ajude!!!

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