Math 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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When I was doing the random quiz, I came across this one and just didn't know what to do. I feel like I know how to do the work fairly well, but on this particular one I couldn't seem to do it.
When it says find the clock time and all there is to go on is the equation, how do you solve it?
I realize this is a quiz and you are not supposed to give me the answers. I will choose another quiz to submit. I just feel stumped on this kind of question. I am so afraid to go up to the college and take a major quiz if I can't even do a simple one like this.
Given the depth vs. clock time function y = f(t) = .018 t^2 + -2.1 t + 78, with depth in cm when clock time is in seconds, find the clock time t such that f(t) = 73.75001 cm and find f(t) when t = 27 sec. Find the clock time when water depth is 43.75001 cm. Using the same function determine the depth at clock time t = 11 sec.
It's fine to ask questions. Do try to give me your best idea what you think the question is asking.
A depth vs. clock time function gives depth as a function of clock time. f(t) = .018 t^2 - 2.1 t + 78 then tells you that f(t) represents the depth and t represents the clock time.
f(t) = .018 t^2 - 2.1 t + 78, so f(t) = 73 simply means .018 t^2 - 2.1 t + 78 = 73. This is a quadratic equation, which I believe you can solve easily by rearranging into standard form and using the quadratic formula.
When t = 27 sec the depth in cm is f(t) = f(27) = .018 * 27^2 - 2.1 * 27 + 78.
Depth is 43 cm when f(t) = 43 cm, i.e., when .018 t^2 - 2.1 t + 78 = 43. Another quadratic equation.
If clock time is t = 11 sec then f(t) = f(11) = ... etc..