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anonymous

  • one year ago

medal-----I just need help figuring out this question not just the answer please The table below represents the velocity of a car as a function of time: Time Velocity (hour) (miles/hours) x y 0 50 1 52 2 54 3 56 Part A: What is the y-intercept of the function, and what does this tell you about the car? Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. Part C: What would be the domain of the function if the velocity of the car was measured until it reached 60

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  1. anonymous
    • one year ago
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    Help please

  2. anonymous
    • one year ago
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    Could you please help me if you can @Nnesha or @whpalmer4 or if you know someone you can tag to help

  3. whpalmer4
    • one year ago
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    What is the definition of the \(y\)-intercept?

  4. anonymous
    • one year ago
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    yes here

  5. whpalmer4
    • one year ago
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    So, what is the definition of the \(y\)-intercept?

  6. anonymous
    • one year ago
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    it is the point on a graph when the x is equalled to 0....?

  7. whpalmer4
    • one year ago
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    Well, it's the value of \(y\) at the point on the graph where \(x=0\). Also the point where the graph crosses the \(y\)-axis.

  8. whpalmer4
    • one year ago
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    So if we graphed the data in the table, would the \(x\)-axis be time, or velocity?

  9. anonymous
    • one year ago
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    x wold be time because the velocity would be y

  10. whpalmer4
    • one year ago
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    good. so now you can tell me the answer to part A...

  11. anonymous
    • one year ago
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    so when x is 0 y is 50

  12. whpalmer4
    • one year ago
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    Yes. And as \(x\) represents time, that's the initial velocity of the car, right? 50 miles/hour

  13. anonymous
    • one year ago
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    yes so all i have to write would be.... the y intercept would be 50 if x is 0 im not sure how to word it

  14. whpalmer4
    • one year ago
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    How about "the y-intercept of the function is 50 miles/hour"?

  15. anonymous
    • one year ago
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    ok ..... :) but it also says what does this tell you about the car? i dont know how to figure that out or does that answer the entire question ? @whpalmer4

  16. whpalmer4
    • one year ago
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    If you look back a few responses, I think you will find that I made a comment about what that represents...

  17. whpalmer4
    • one year ago
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    The data in the table shows the car starting at 50 mph and going faster with each passing hour, right?

  18. anonymous
    • one year ago
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    yes by 2 miles per hour

  19. whpalmer4
    • one year ago
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    no, but close: it changes by 2 miles per hour per hour...

  20. whpalmer4
    • one year ago
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    or miles/hour^2

  21. anonymous
    • one year ago
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    yes :) your right

  22. whpalmer4
    • one year ago
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    no, "you're right" :-)

  23. whpalmer4
    • one year ago
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    On to part B! How do you find the average rate of change?

  24. anonymous
    • one year ago
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    average rate of change for an function f(x) is intervals from a to b so the formula is f(b)-f(a) over b-a

  25. whpalmer4
    • one year ago
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    okay, do you understand how to translate your table values into a,b,f()?

  26. anonymous
    • one year ago
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    @whpalmer4 i was reading my notes from class would F(a) be x and f (b) be y

  27. anonymous
    • one year ago
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    so would it be from the table 52-50 over 1-0 ??? not sure

  28. whpalmer4
    • one year ago
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    Well, \(x\) takes on the values of \(a\) and \(b\) and \(y = f(x)\) So at \(x=1 \text{ hour}\) we have \(a = x = 1\text{ hour}\) and \(f(1) = \) what?

  29. anonymous
    • one year ago
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    52

  30. whpalmer4
    • one year ago
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    Yes. so \(a = 1\) and \(f(a) = 52\) Now how about \(b\) and \(f(b)\)?

  31. anonymous
    • one year ago
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    b =2 and f(b)=2

  32. anonymous
    • one year ago
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    f(b) =54 oops

  33. whpalmer4
    • one year ago
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    why does b = 2? and f(b) is the value of the function at x=b=2, which doesn't appear to be 2 when I read the table... Here's the initial problem again: Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. ^^^^^^^^^

  34. whpalmer4
    • one year ago
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    well, my attempt to underline "\(x=3\)" didn't work so well!

  35. anonymous
    • one year ago
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    ohhh So b =3 and f(b)=56

  36. whpalmer4
    • one year ago
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    yes! we are trying to find the average rate of change over that 2 hour span, so \(a\) is the value of \(x\) or \(t\) at the start, and \(b\) is the value of \(x\) or \(t\) at the end, and \(f(a)\) and \(f(b)\) are the values from the table.

  37. whpalmer4
    • one year ago
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    So our average rate of change between \(a\) and \(b\) is as your formula states: \[\text{avg rate of change} = \frac{f(b)-f(a)}{b-a}\]

  38. whpalmer4
    • one year ago
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    Can you plug in the values?

  39. anonymous
    • one year ago
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    56-52/ 3-1=4/2 =2

  40. whpalmer4
    • one year ago
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    Okay, if you are going to write equations on one line like that, you MUST use parentheses appropriately to show what isn't being shown by position! what you wrote is actually equal to this: \[56 - \frac{52}{3} - 1 = \frac{4}{2}\] I know that isn't what you meant. (56-52)/(3-1) = 4/2 would unambiguously convey the intended meaning

  41. whpalmer4
    • one year ago
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    What will the units for your answer be? miles? miles per hour? something else?

  42. anonymous
    • one year ago
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    oh ok thanks for that tip and my units would be miles per hour or miles/hour

  43. anonymous
    • one year ago
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    so ithe change of rate would be 2 miles per hour

  44. whpalmer4
    • one year ago
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    right number, wrong unit \[\frac{56 \text{ miles/hr} - 52 \text{ miles/hr}}{3\text{ hr} - 1\text{ hr}} = \frac{4\text{ miles/hr}}{2\text{ hr}} = 2 \text{miles/hr}^2\] \]

  45. anonymous
    • one year ago
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    can you explain why is it hr^2

  46. anonymous
    • one year ago
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    I dont know how to get the little 2 sorry

  47. whpalmer4
    • one year ago
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    right? if we continue accelerating at this rate, for every hour that passes, we will increase our speed by 2 miles/hour. If we accelerate for 10 hours, we will add \[10\text{ hours}*2\frac{\text{ miles}}{\text{hour*hour}} = 10\cancel{\text{ hours}}*2\frac{\text{ miles}}{\text{hour}*\cancel{\text{hour}}} \] \[= 20 \frac{\text{ miles}}{\text{hour}}\] to our speed

  48. anonymous
    • one year ago
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    ohhh yes ........ :0

  49. anonymous
    • one year ago
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    thanks for explaining it

  50. anonymous
    • one year ago
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    my mom says we have to go to church can i see if your on later to finish :)

  51. whpalmer4
    • one year ago
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    the way it "builds up" in physics terminology: \[x = v t\]distance = speed * time \[v = a t\]velocity = acceleration * time

  52. whpalmer4
    • one year ago
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    Sure, say hi to the big guy for me :-)

  53. anonymous
    • one year ago
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    lol i will thank you so much

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