Steve drove at a constant rate to the beach for a vacation. In the equation below, t is the time in hours it took Steve to drive to the beach. 66t=330 What is the unit rate in the equation above? A. 5 miles per hour B. 264 miles per hour C. 66 miles per hour D. 330 miles per hour
Josh: Please show how you arrive at your answers. In the long run you'd get better feedback by doing so.
i only post the ones i have no clue on
sorry i was on brb
Are you sure that the original problem doesn't provide units of measurement with the "66" and "330?"
no i'm saying i DON'T KNOW
i'll check that out
i still don't get it
not all of us are good at math xD
One o the more common applications of algebra is the relationship Distance = Rate * Time. Here, "rate" refers to speed. d = r * t.
Given d = r * t, solve for the rate, r.
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what is r 330?
ok i see it's speed
but i don't see where they give that is it the 330?
Gave you an algebraic formula ( d = r*t ) and asked you to solve it algebraically for the rate (speed), r. Given d = rt, solve for r, please.
This was your original post: Steve drove at a constant rate to the beach for a vacation. In the equation below, t is the time in hours it took Steve to drive to the beach. 66t=330
Given d = r * t, solve for r.
yes but if ir's r * t how do i find r?
r anyway
Solve for r algebraically. That means: Isolate r. If d = r * t, divide both sides of the equation by t. That will leave r alone on the right side.
i still don't get it i'm right where i started
d = r * t. distance equals rate times time. Solve for the rate, r, by dividing both sides of this equation by t: d r* t ---- = ----- t t Cancel out the "t" on the right side, please. Your result?
rate = t 330=66t since you have to make t the subject (t alone on lhs) divide 66 on both sides t = 330/66 t=.....?
but if t is the distance then what's d?
5
Faiq: Joshua needs to solve for the RATE, not for the time.
i have NO idea what you're talking about at all :l
Have you not done any preparation? Read an introductory algebra book? It's expecting a lot, perhaps too much, if you come to OpenStudy to expect people to teach you concepts from the very beginning.
forget it ima guess onto the next question for me
oh sorry t = speed correction
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