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anonymous

  • 5 years ago

can anybody help me?? 1) Simon’s cell phone plan charges a $16 monthly service fee plus $0.04 per minute. Megan’s plan charges $0.12 per minute with no monthly fee. How many minutes do each of them have to use to pay the same amount? 1A) what the intersection represents 1B) coordinates of intersection

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  1. anonymous
    • 5 years ago
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    Hello, I would love to help you with this problem. Have you started this problem already or do you have an idea of what the steps should be?

  2. anonymous
    • 5 years ago
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    hi....I'm all lost... :( ...please help me from the first

  3. anonymous
    • 5 years ago
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    alrighty then. I think the first step is to set up a system of equations. Have you studied this before?

  4. anonymous
    • 5 years ago
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    :( .....noooooo .... need to start with x and y ?????

  5. anonymous
    • 5 years ago
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    Well that is what we will do then. Why don't we make the first equation with x and y look something like this: \[.04x+16=y\] this equation tell us how much Simon will spend each month for his cellphone. "Y" tells us what his total monthly charge is. The .04x is part of the fee where "X" is how many minutes he uses each month and the 16 is the $16 fee he is charged automatically. Does this make sense?

  6. anonymous
    • 5 years ago
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    ohh.... is it? .....and like this I can say.: .12x = y......right?

  7. anonymous
    • 5 years ago
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    you there?

  8. anonymous
    • 5 years ago
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    :(

  9. anonymous
    • 5 years ago
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    Yes that is right. So your second equation is going to be \[.12x=y\] and that is going to tell us how much megan's bill will be. where again "x" is the number of minutes she used that month and "y" is the bill total.

  10. anonymous
    • 5 years ago
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    but I wonder what would be the next??

  11. anonymous
    • 5 years ago
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    Are you righting this down? Because we are about to go back to the original question with this new information and it will be helpful to keep track of our steps :)

  12. anonymous
    • 5 years ago
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    yeah.....I'm all set to go ahead... :)

  13. anonymous
    • 5 years ago
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    okay so look at the two equations you have on your paper. Do you see what they have in common? They both use the same variables, which are "x" and "y".

  14. anonymous
    • 5 years ago
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    yeah

  15. anonymous
    • 5 years ago
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    Well the first part of the question is asking us for how many minutes they must use to have the same bill. So we know that if "x" is the number of minutes they use and "y" is their bill, these values must be the same for both equations right?

  16. anonymous
    • 5 years ago
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    ohh....yeah...

  17. anonymous
    • 5 years ago
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    okay well why don't we set the two equations equal to each other? What I mean by equal to each other is use what they have in common: \[.04x+16=y | y=.12x\] see where you could set them equal to each other? try putting the two together like this: \[.04x+16=.12x\] they have the same answer for "y" so what we can do is take advantage of this by setting the two equations to be "equal" to each other.

  18. anonymous
    • 5 years ago
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    right... :) ....and I"ve solved it....the value of x should be 200

  19. anonymous
    • 5 years ago
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    but what about the next parts??? :(

  20. anonymous
    • 5 years ago
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    YES! So if x=200 and thats the number of minutes they used then what we really have is 1/2 the co-ordinates to graph this problem. Why don't we take our newfound value of "x" and plug it into each equation separately? so we can have: \[.04(200)+16=y \] and \[.12(200)=y \] and then solve for y in each one of the equations to get Simon and Megan's bill totals.

  21. anonymous
    • 5 years ago
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    and y = 24

  22. anonymous
    • 5 years ago
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    yup it is! Now you have x and y values for both equations. But they are the same... how does this help? Well look at the question 1A). What does the intersection represent? Well consider that if 2 equations have the same point in common they are intersecting. What this means for us is that at the point of intersection, megan and simon used the same number of minutes and had the same bill. (remember that they both used 200 minutes and both of them had the same bill amount of $24.00)

  23. anonymous
    • 5 years ago
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    I'm trying to understand all...please be with me

  24. anonymous
    • 5 years ago
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    its alright take your time ;) is there a specific part you're having trouble with?

  25. anonymous
    • 5 years ago
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    you mean the point of intersection would be (x,y) or (200, 24) ??

  26. anonymous
    • 5 years ago
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    THAT IS CORRECT! If you were to draw a graph of the functions on the x,y plane they would intersect at (200,24) and what that means or represents is that when they have both used 200 minutes their phone bills will be the same.

  27. anonymous
    • 5 years ago
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    okkkkkkkkkkkkkkk...........that's great.........thanks a lottttttttttt ... :)

  28. anonymous
    • 5 years ago
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    can you help me with one more problem???

  29. anonymous
    • 5 years ago
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    yup I sure can :)

  30. anonymous
    • 5 years ago
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    :) ....so kind of you......

  31. anonymous
    • 5 years ago
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    look at the sequence of numbers in the set: {7^1, 7^2, 7^3, 7^4.... 7^1998} How many of the numbers in the set have a ones digit of 9?

  32. anonymous
    • 5 years ago
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    Hahaha I'm really really really bad with sequences but I will give it a shot. {7^1=7,7^2=49,7^3=343,7^4=2401,7^5=16807...7^1998=a number so big it won't show up on most calculators but mine says it ends in 9....} I think we need to find a pattern to the system. So our first interaction with a 9 in the one's place is at the value of 7 squared or 7^2. May I ask what class is this for? I am gonna guess an algebra course or something similar? ;)

  33. anonymous
    • 5 years ago
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    :( .....I"m in grade eight ....I've got hint like ... There are 1,998 numbers in the set. And the terms are: 7^1 = 7 7^2 = 49 7^3 = 343 7^4 = 2401 7^5 = 16807 7^6 = 117649 7^7 = 823543 7^8 = 5764801 ... The ones digit is the very last number. Note that the ones digit has a repeating pattern: 7, 9, 3, 1, 7, 9, 3, 1, ..., 9 and the pattern is four terms long.

  34. anonymous
    • 5 years ago
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    Now... The ones digit is a 9 when the exponent of 7 is (4n-2), where n is a positive integer:

  35. anonymous
    • 5 years ago
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    and n would be 2,6,10,......but...I need to solve what could be the last value of n for which the ones digit will be 9....... and I'm lost

  36. anonymous
    • 5 years ago
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    :(

  37. anonymous
    • 5 years ago
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    okay lets see....

  38. anonymous
    • 5 years ago
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    to start I think I would take note of the fact that the 9 shows up in the one's place every "4" different exponent expansions... So with that prediction lets guess and see if 7^10 will end in a 9 (or have a 9 in the one's place). The answer I received with my calculator program is: 282475249 which does end with a 9 in the ones place. So if every 4 terms a 9 will show up we can calculate the total number. the equation they give you of (4n-2) is right because our first encounter with a 9 in the ones place happened at 7^2 which is only 2 terms into the sequence. Thats where the 2 in (4n-2) comes from. They didn't tell you what to set (4n-2) equal to directly but we are supposed to set it equal to 1998. :) \[4n-2=1998\] or more accurately \[4n=2000\] since we will move 2 over by adding it :)

  39. anonymous
    • 5 years ago
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    but is the 1998th term has a 9 in the one's place?

  40. anonymous
    • 5 years ago
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    so the question is asking whether or not 7^1998 has a 9 in the ones place?

  41. anonymous
    • 5 years ago
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    no no...I"m asking it....does it contain a 9 in the one's place? or what is the nearest number of 1998 containing9 in one's place?

  42. anonymous
    • 5 years ago
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    ahhhh okay gimme just a second I think I've got it now ;)

  43. anonymous
    • 5 years ago
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    okay I am going to go back to your original question. "look at the sequence of numbers in the set: {7^1, 7^2, 7^3, 7^4.... 7^1998} How many of the numbers in the set have a ones digit of 9? " is that the way the question actually looks or what it says? If not what does the question state? If its really looking for a number containing a 9 in the ones place that follows the 7^(4n-2) pattern then it would be 49...

  44. anonymous
    • 5 years ago
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    no....it's asking how many numbers are there in the set: {7^1, 7^2, 7^3, 7^4.... 7^1998} containing 9 in the one's place

  45. anonymous
    • 5 years ago
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    oh well then we kinda already answered that ;) that was using the equations we were looking at earlier. 4n-2=1998. If there are 1998 numbers in this set. meaning exponents of 7 from 1 to 1998 how many of them will have a 9 in the ones place. Well lets follow our equation. 4n <- because every 4th term will have a 9 in the ones place. and then -2 <- because the first time we see a 9 in the sequence is the 2nd term. So our equation states that the 2nd term and every fourth term after that will have a 9 in the ones place. Then 1998 is the last exponent we are going to evaluate. So when we solve 4n-2=1998 for n we will see that there are 500 numbers in this set that will end with a 9 in the ones place.

  46. anonymous
    • 5 years ago
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    that means that there are 500 numbers where 7^some number will have a value where 9 is in the one's place

  47. anonymous
    • 5 years ago
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    ohhh.......ok....now got it.....

  48. anonymous
    • 5 years ago
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    yeah sorry for the confusion. I am really bad with number sets and patterns. :) Do you have any other questions?

  49. anonymous
    • 5 years ago
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    ohh no....it helped me and you are really good........ :) ...but yeah....one last question.... :(

  50. anonymous
    • 5 years ago
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    haha thank you for the compliment. And what is the question?

  51. anonymous
    • 5 years ago
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    How many solutions does 3 cos2 x = 1 have on the interval [0, 2π)

  52. anonymous
    • 5 years ago
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    I saw that question float by earlier. I'll give it a shot too ;)

  53. anonymous
    • 5 years ago
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    okay what do you know about the interval of [0,2pi]?

  54. anonymous
    • 5 years ago
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    0 to 360 degree interval

  55. anonymous
    • 5 years ago
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    correct. Have you ever done any graphing using [0,2pi) before? Solving this problem graphically is the quickest and easiest way but there are other methods if you have not studied graphing of sine and cosine functions.

  56. anonymous
    • 5 years ago
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    please explain me first and then I"ll try to do it graphically by myself... :)

  57. anonymous
    • 5 years ago
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    you there??.... :(

  58. anonymous
    • 5 years ago
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    haha okay. Well the basic graph of a cosine wave oscillates (moves back and forth) vertically from -1 and 1 on a graph. This means that the y-value of the graph is going to change from -1 to 1 over and over again. Have a look at this graphic I made to explain what it is. http://i625.photobucket.com/albums/tt338/npsgaming/Math%20Help%20On%20OPEN%20STUDY/CosineWave.jpg?t=1301026735

  59. anonymous
    • 5 years ago
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    yeah...that's clear to me

  60. anonymous
    • 5 years ago
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    So looking at that graph you can see the y value is changing from -1 to 1 as it moves along the x-axis. On the x-axis you will see pi and 2pi labeled. This will help us in a moment. With this visual in mind, think about the interval of [0,2pi) it includes the value 0 and and doesn't include 2pi but everything between there is a valid answer.

  61. anonymous
    • 5 years ago
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    So we will use this graph again when it comes time to plot our function 3cos(2x)=1. The interval limits our solutions to anytime the line crosses the x-axis. Here is another image that is the graph of 3cos(2x)=1. http://i625.photobucket.com/albums/tt338/npsgaming/Math%20Help%20On%20OPEN%20STUDY/3cos2xeq1.jpg?t=1301027056 that is what the graph of 3cos(2x)=1 looks like. Now take not of where 0 and 2pi are. How many times do you see the line crossing the x-axis? (its the red lines :) )

  62. anonymous
    • 5 years ago
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    6 times

  63. anonymous
    • 5 years ago
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    thats how many times are shown, but remember the interval of 0 to 2pi limits our solutions to only the lines that are between 0 and 2pi. So while there are 6 lines shown, only 4 are within the interval of [0,2pi)

  64. anonymous
    • 5 years ago
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    hmmmm.....but how can I do it without graph??

  65. anonymous
    • 5 years ago
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    have you ever used the unit circle to answer a problem? (thats where the 360 degrees you mentioned earlier comes from)

  66. anonymous
    • 5 years ago
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    not sure....can you show me the steps so that I can have an idea???

  67. anonymous
    • 5 years ago
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    :(

  68. anonymous
    • 5 years ago
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    haha I will do my best. I was never any good at geometry either ;)

  69. anonymous
    • 5 years ago
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    does the problem ask you for the specific solutions? or is it just asking for how many? :)

  70. anonymous
    • 5 years ago
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    cuz I can show you using the unit circle and triangle how many solutions there are :)

  71. anonymous
    • 5 years ago
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    it's just asking how many solutions

  72. anonymous
    • 5 years ago
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    okay

  73. anonymous
    • 5 years ago
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    this link will show you the unit circle. Notice that its radius is 1. and the unit circle is broken up into quadrants just like a graph. There are 4 quadrants. 0-90deg, 90-180deg,180-270deg, and 270-360deg. http://www.processing.org/learning/trig/imgs/unit_circle.jpg Does this make sense so far? (this is a pretty difficult topic and I'm not sure I even knew what the unit circle was until 9th grade or so.

  74. anonymous
    • 5 years ago
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    yeah ...got it....but what about the problem???? ... :(

  75. anonymous
    • 5 years ago
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    okay so our problem 3cos(2x)=1 has 4 possible solutions. Because our interval is from 0 to 2pi this includes all 4 quadrants of the unit circle.... Each quadrant has an equivalent version of the answer. Right now im working on trying to get the exact answers.... bear with me ;) Its been a few years since I touched the unit circle ;)

  76. anonymous
    • 5 years ago
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    :).... you have already helped me so much.........I'll be happy to wait.

  77. anonymous
    • 5 years ago
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    have you ever heard of the arc cosine? (sometimes written as: \[\cos^{-1} \]

  78. anonymous
    • 5 years ago
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    yeah..

  79. anonymous
    • 5 years ago
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    okay I think I may have it then ;)

  80. anonymous
    • 5 years ago
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    one solution is \[1/2 \cos^-1(1/3)\] that is for quadrant 1 and you can get that by using some algebra. 3cos(2x)=1 into cos(2x)=1 but from there I'm no sure where to go :(

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