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anonymous

  • 5 years ago

what exponential function passes thru (0,4) and (-2,64)

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  1. mattfeury
    • 5 years ago
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    the basic form for an exponential equation is \[y = ax^2 + b\] so I'd plug in (0,4) and we can solve for b then plug in (-2,64) to solve for a

  2. mattfeury
    • 5 years ago
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    \[4 = 0*0^2 + b\]\[b = 4\]\[64 = a*(-2)^2 + 4\]\[15 = a\] boom. plug those in your general equation (above) and you've got an answer!

  3. anonymous
    • 5 years ago
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    these are the choices tho 1) f(x)=4(1\4)^x 2) f(x)=4^x 3)f(x)=4(1\4)^-x 4)f(x)=-4(1\4)^x

  4. anonymous
    • 5 years ago
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    thats why im confused?

  5. mattfeury
    • 5 years ago
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    well you can plug each set of numbers into all those functions and see which works. it looks like (1) will work for both sets.

  6. mattfeury
    • 5 years ago
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    ahh. i misread 'exponential' to be 'quadratic'. My answer is one example, but there are many functions that would work.

  7. anonymous
    • 5 years ago
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    thank you so much and by any chance do you know this one: 3 sides of a rectangular garden are enclosed using 120 meters of fencing. what is the maximum area of garden?

  8. mattfeury
    • 5 years ago
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    hmm what level math are you in? I am tempted to use derivatives, but not if you haven't learned them yet.

  9. anonymous
    • 5 years ago
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    algebra 2

  10. anonymous
    • 5 years ago
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    and yeah i havent learned that yet lol

  11. mattfeury
    • 5 years ago
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    well I'm thinking that the maximum area would be a perfect square rather than a rectangle, don't you think? this would mean each of the three sides are equal.

  12. anonymous
    • 5 years ago
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    The basic form is\[y=a^{bx}\]

  13. myininaya
    • 5 years ago
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    ok we have x+x+y=120 so y=120-2x so A=xy=x(120-2x)=120x-2x^2 so if we find the vertex of this parabola we can find the max

  14. anonymous
    • 5 years ago
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    I get \[y=2^{-3x}\]

  15. myininaya
    • 5 years ago
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    i got y=4^(-3x/2) for that one

  16. anonymous
    • 5 years ago
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    Actually, I don't...!

  17. anonymous
    • 5 years ago
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    so with the choices 1) 3,600m^2 2) 1,800m^2 3)1,600m^2 4)900m^2 whats the correct answer

  18. myininaya
    • 5 years ago
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    anyways the vertex of that parabola is (0,120) and you find that by writing the parabole as a*(x-h)^2+k where (h,k) is the vertex

  19. myininaya
    • 5 years ago
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    oops one sec i made a mistake

  20. anonymous
    • 5 years ago
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    \[y=4e^{-x \ln4}\]

  21. anonymous
    • 5 years ago
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    If you let y be length of one of the sides, and x be the length of the other two sides, then y+2x=120. The area is given by A=xy, so since y=120-2x, just substitute for y in A=xy. This gives A=x(120-2x). You are trying to find x that maximizes this. Since the graph of this will be an upside down parabola with roots at x=0 and x=60, this means the maximum will occur at x=30 (halfway between the roots). So the max area is A=30*(120-2*30)=30*60=1800.

  22. anonymous
    • 5 years ago
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    When x is 0, y is 4. When x is -2, y is 64.

  23. myininaya
    • 5 years ago
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    the vertex of the parabola is (30,1800) since A=-2(x-30)^2+1800

  24. anonymous
    • 5 years ago
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    Why are we talking about quadratics?

  25. anonymous
    • 5 years ago
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    oh okay so the answe is 2

  26. anonymous
    • 5 years ago
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    1800

  27. anonymous
    • 5 years ago
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    \[y=ae^{bx}\]When x=0, y=4, so\[4=ae^0=a\]so a=4. When x=-2, y=64, so\[64=4e^{-2b} \rightarrow 16=e^{-2b} \rightarrow \ln 16 = -2b \rightarrow b=-\frac{1}{2}\ln 16=-\ln 4\]so b = -ln(4). hence your equation is,\[y=4e^{-(\ln 4)x }\]

  28. myininaya
    • 5 years ago
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    you're so smart lokisan lol

  29. anonymous
    • 5 years ago
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    haha...but wait, there's more...

  30. anonymous
    • 5 years ago
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    \[e^{-(\ln4)x}=e^{\ln \frac{1}{4}x}=\left( e^{\ln \frac{1}{4}} \right)^x=\frac{1}{4^x}\]so\[y=4e^{-(\ln 4)x}=4.\frac{1}{4^x}=\frac{1}{4^{x-1}}=4^{1-x}\]

  31. anonymous
    • 5 years ago
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    So your equation can be written as,\[y=4^{1-x}\]

  32. anonymous
    • 5 years ago
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    Silence...

  33. anonymous
    • 5 years ago
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    lol u guys are so good at this im so confused. but okay if thats my equation whats the answer, 1 2 3 or 4

  34. anonymous
    • 5 years ago
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    3600 or 1800 or 1600 or 900

  35. anonymous
    • 5 years ago
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    What do you actually need to find out? I thought you just had to come up with the equation.

  36. anonymous
    • 5 years ago
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    imrickjames is here...haven't seen you in a while ;]

  37. anonymous
    • 5 years ago
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    noo .. i wrote up there earlier the choices...its asking me what the maximum area of the garden is

  38. anonymous
    • 5 years ago
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    Okay...let me check the multiple choices...

  39. anonymous
    • 5 years ago
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    What garden? Isn't that another question?

  40. anonymous
    • 5 years ago
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    Oh, I see the garden question. So the others didn't work it out?

  41. anonymous
    • 5 years ago
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    no i just need to answer the garden question. i need to choose one of those 4 and show work.

  42. anonymous
    • 5 years ago
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    1800

  43. anonymous
    • 5 years ago
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  44. anonymous
    • 5 years ago
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    Thanks for fanning me, Matt :)

  45. anonymous
    • 5 years ago
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    thank u ..wat about if a car is purchased for 20,000. the value of the car decreases by 10% each year. about how many years will it take for the vallue of the car to reach 8,000

  46. mattfeury
    • 5 years ago
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    hey! you can't know it was me! heh :). impressed by your scan.

  47. anonymous
    • 5 years ago
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    About 8.7 years.

  48. anonymous
    • 5 years ago
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    Impressed by the scan...so *that's* how I increase the fan base ;)

  49. anonymous
    • 5 years ago
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    how did u get that thoo

  50. anonymous
    • 5 years ago
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    These types of equations with growth and decay in proportion to themselves follow the form\[p(t)=p(0)e^{kt}\]where p(0) is the initial value and k some constant. For growth, it's positive, for decay, negative. You get this equation by solving the differential equation,\[\frac{dp}{dt}=kp\]which is separable (I don't know what level you're at). Anyway, with this equation, you use your data to find out what p(0) and k are, then answer the question.

  51. anonymous
    • 5 years ago
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    Here, p(0) is 20,000. To find k, you know that after 1 year, the car's value is 10% less, so it's value at time t=1 is 18,000. So,\[18,000=20,000e^{k}\]Divide both sides by 20,000 and take the natural log to find k. Now you have your p(0) and your k, so you can find p(t) at any time t.

  52. anonymous
    • 5 years ago
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    okay thank uuu ! and if i have to find |3+i| that equals what ?

  53. anonymous
    • 5 years ago
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    You want to find t when p(t) is 8000, so you set it up as,\[8000=20000e^{ \ln (\frac{9}{10})t}\]You have to solve for t. Divide both sides by 20,000 and again, take the natural log to find\[\frac{8000}{20000}=\ln (\frac{9}{10})t \rightarrow t =\frac{8000/20000}{\ln (9/10)} \approx 8.7yrs\]

  54. anonymous
    • 5 years ago
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    It's asking for the magnitude. The magnitude of a complex number is z is,\[|z|=|x+iy|=\sqrt{x^2+y^2}\]You consider the real and imaginary part (note, the imaginary part is NOT iy, it's y, and y is real...yi is imaginary).

  55. anonymous
    • 5 years ago
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    \[|3+i|=\sqrt{3^2+1^2}=\sqrt{10}\]

  56. anonymous
    • 5 years ago
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    Good luck. If you have anymore questions, post them separately...I need to log out :D

  57. anonymous
    • 5 years ago
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    thank youuuuu so muchhh

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