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Dominik [7]
3 years ago
13

Tobias sent a chain letter to his friends, asking them to forward the letter to more friends. The number of people who receive t

he email increases by a factor of 4 every
9.1 weeks, and can be modeled by a function, P, which depends on the amount of time, t (in weeks).
Tobias initially sent the chain letter to 37 friends.

Write a function that models the number of people who receive the email t weeks since Tobias initially sent the chain letter.

P(t)=
Mathematics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

        P(t)=37\cdot 4^{t/9.1}

Explanation:

This is an exponential growing: there is a constant growing<em> factor </em>which multiplies the value number of letters sent every week.

The growing factor <em>4 letters every 9.1 weeks</em> means that every 9.1 weeks the number of letters is multiplied by 4.

Then, if the number of weeks is t, the number of times the number of letters increase is t/9.1.

Then, the exponential<em> function</em> <em>that models the number of people who receive the email t weeks since Tobias initially sent the chain letter</em>, has the form:

        P(t)=Initial\text{ }value\cdot 4^{(t/9.1)}

<em>Tobias initially sent the chain letter to 37 friends</em>; thus, the initial value is 37, and the complete function is:

        P(t)=37\cdot 4^{t/9.1}, where t is the number of weeks since Tobias initially sent the chain letter.

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Answer:

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Step-by-step explanation:

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∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

 xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

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∵ We have term xy that means we rotated the graph about

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∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

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∴ x' = x/√2 - y/√2 = (x - y)/√2

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* Lets substitute x' and y' in the 1st answer

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∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

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6 0
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