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DiKsa [7]
3 years ago
5

After every eighth visit to a restaurant you receive a free beverage. After every tenth visit you receive a free appetizer. If y

ou visit the restaurant 100 times, on which visits will you receive a free beverage and a free appetizer? At which visit will you first receive a free beverage and a free appetizer? Drag and drop the correct value into each box to complete the statement. You will receive a free beverage and a free appetizer on the and visits, and you will first receive a free beverage and a free appetizer on the visit.
Mathematics
2 answers:
AleksandrR [38]3 years ago
8 0

Answer:

Step-by-step explanation:

Kaylis [27]3 years ago
4 0

Answer:

40th and 80th visit

the first visit where you receive both is the 40th

Step-by-step explanation:

personally, I believe that the simplest way to solve this is use an excel spreadsheet to see which visits would include both a free beverage and a free appetizer:

  • both rewards would happen only twice, at the 40th and 80th visit.

you can also calculate it mathematically since multiples of 8 and 10 do not repeat themselves very much, only when 8 is multiplied by 5 and 10.

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It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

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\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

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