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Doss [256]
2 years ago
13

A person is trying to quit chewing gum. They decide they will have 64 pieces in week 1, but each week have half as many as the w

eek before. The function f represents the number of pieces of gum they will have in week n. What is a reasonable domain for this function? Explain your reasoning.
Mathematics
1 answer:
blagie [28]2 years ago
4 0

Function is missing.

It is;

f(n) = ½⋅f(n − 1) for n ≥ 2

Answer:

Domain for this function is;

n ≥ 1

Step-by-step explanation:

We are told They decide they will have 64 pieces in week 1,

Thus;

f(1) = 64

From the function given;

f(n) = ½⋅f(n − 1)

So;

f(2) = ½•f(2 - 1)

f(2) = ½•f(1)

f(2) = ½ • 64

f(2) = 32

Also;

f(3) = ½•f(3 - 1)

f(3) = ½•f(2)

f(3) = ½ × 32 = 16

Looking at the steps, they follow a general pattern of;

f(n) = 64•½^(n - 1) for n ≥ 1

Thus, the domain for the function for n weeks after the first one is n ≥ 1

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The divisor x + 3 is a polynomial, also about x, but with degree 1.

By the division algorithm, the quotient should be of degree 3 - 1 = 2, while the remainder shall be of degree 1 - 1 = 0 (i.e., the remainder would be a constant.) Let the quotient be a\,x^2 + b\, x + c with coefficients a, b, and c.

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Start by finding the first coefficient of the quotient.

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Now, given that a = 4, rewrite the right-hand side:

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The term with a degree of two on the left-hand side has coefficient (-12). Since the only term on the right hand side with degree two would have coefficient b, b = -12.

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Subtract -12x^2 -36x from both sides of the equation:

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By the same logic, c = 38.

Hence the quotient would be (4x^2 - 12x + 38).

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