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A 15-inch candle is lit and burns at a constant rate of 1.2 inches per hour. Let t represent the number of hours since the candle was lit, and uppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit.
a. Write a function formula for f f(t)- [Preview] syntax error #
b. What is the domain of f relative to this contex? Enter your answer as an interval. # Preview, no answer given
c. What is the range of f relative to this context? Enter your answer as an interval *Preview no answer given
d. Solve f(t) 3.6 for it. 算 Preview
e. What does your solution in part (d) represent in this context? Select all that apply. The length of the candle (in inches) 3.6 hours after it was lit. How many bhous since the candle was lir wen i is36 mches loe OHow long it takes for the candle to burn out.

Sagot :

Answer:

[tex]f(t) = 15 -1.2t[/tex]

[tex]0 \le t \le 12.5[/tex]

[tex]15 \ge f(t)t \ge 0[/tex]

The length of the candle (in inches) 3.6 hours after it was lit is 0.95 inches

Step-by-step explanation:

Given

[tex]a = 15[/tex] ---- candle length

[tex]r =- 1.2[/tex] --- rate (it is negative because the candle length reduces)

Solving (a): The function, f(t)

This is calculated as:

[tex]f(t) = a + rt[/tex]

So, we have:

[tex]f(t) = 15 + -1.2 * t[/tex]

[tex]f(t) = 15 -1.2t[/tex]

Solving (b): The domain

The domain, in this case, represents time from 0 till the candle burns out.

Set [tex]f(t) = 0[/tex] to calculate the time the candle burns out

[tex]f(t) = 15 -1.2t[/tex]

[tex]0 = 15 - 1.2t[/tex]

Collect like terms.

[tex]1.2t = 15[/tex]

Solve for t

[tex]t = 15/1,2[/tex]

[tex]t = 12.5[/tex]

Hence, the domain is: [tex]0 \le t \le 12.5[/tex]

Solving (c): The range

The domain, in this case, represents the candle height from 15 till the candle burns out (0 inches)

Hence, the range is: [tex]15 \ge f(t)t \ge 0[/tex]

Solving (d): [tex]f(t) = 3.6[/tex]

Recall that:[tex]f(t) = 15 -1.2t[/tex]

Sp, we have

[tex]15 - 1.2t - 3.6[/tex]

Collect like terms

[tex]12t =15 -3.6[/tex]

[tex]12t =11.4[/tex]

Divide by 12

[tex]t =0.95[/tex]