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2 years ago

A function f gives the number of stray cats in a town t years since the town started

Mathematics

1 answer:

Bingel [31]2 years ago

6 0

Answer:

f(0) = 243 . . . cats at the start

Step-by-step explanation:

The exponential function tells you the number of cats after t years is ...

f(t) = 243(1/3)^t

When t=0, the exponential factor is 1, so the value of the function is the multiplying constant.

f(0) = 243

This means there were 243 stray cats in town when the animal control program started.

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What is f(5) if f(1) = 3. 2 and f(x 1) = Five-halves(f(x))?

stealth61 [152]

Function assign value from one set to another. The value of f(5), if f(1) = 3.2 and f(x+1) = Five-halves(f(x)) is 125.

<h3>What is Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

As it is given the value of the function f(1) is 3.2, while the value of f(x+1) is $f(x+1) = \dfrac{5}{2}[f(x)]$ , therefore, in order to find the value of f(5), we need to calculate the value of f(4).

f(2)

$f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(2)=f(1+1) = \dfrac{5}{2}[f(1)]\\\\f(2) = 2.5 \times 3.2\\\\f(2) = 8$

f(3)

$f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(3)=f(2+1) = \dfrac{5}{2}[f(2)]\\\\f(3) = 2.5 \times 8\\\\f(3) = 20$

f(4)

$f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(4)=f(3+1) = \dfrac{5}{2}[f(3)]\\\\f(4) = 2.5 \times 20\\\\f(4) = 50$

f(5)

$f(x+1) = \dfrac{5}{2}[f(x)]\\\\f(5)=f(4+1) = \dfrac{5}{2}[f(4)]\\\\f(5) = 2.5 \times 50\\\\f(5) = 125$

Hence, the value of f(5), if f(1) = 3. 2 and f(x+1) = Five-halves(f(x)) is 125.

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