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Curry

  • 4 years ago

hard question please help me ABC is a triangle with sides AB = 6m, BC =8m, and AC = 10m. A line k in the plane of the triangle ABC moves along the segment AC at the rate of 1cm per sec. The line starts at A and ends at C and is always perpendicular to AC. 1. How long does it take the line to reach the point B? 2. How long does it take the line to bisect the area of the triangle? 3. What is the area of the region that the line sweeps in its movement after 6min 45 sec? 4.What is the area of the region that the line sweeps at any give moment from the start of its movement?

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  1. Curry
    • 4 years ago
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    i need help on 2/3/4

  2. anonymous
    • 4 years ago
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    if you can, draw the figure to go along with it.

  3. Curry
    • 4 years ago
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    |dw:1328077516987:dw|

  4. anonymous
    • 4 years ago
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    this IS a tough one...

  5. Curry
    • 4 years ago
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    yep it was a precalc problem. i solved 1,3,4 and i need 2 now but im not sure if my answer for 3 and 4 are right but i am sure of number 1

  6. anonymous
    • 4 years ago
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    the key is in part 4. if you can figure that out, you should be able to do the rest very easily. give me a moment.

  7. Curry
    • 4 years ago
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    i got 2/3cm^2 for every 1 centimeter it moves

  8. Curry
    • 4 years ago
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    but i t

  9. Curry
    • 4 years ago
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    but i t

  10. Curry
    • 4 years ago
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    it is wrong cause there is no way that is possible

  11. anonymous
    • 4 years ago
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    sorry, had to find some paper

  12. anonymous
    • 4 years ago
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    i think you have it right, actually. you have a basic 3/4/5 triangle here. when the line k is moving across, it is creating a smaller 3/4/5 triangle because it is perpendicular to AC.

  13. anonymous
    • 4 years ago
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    so the area of the small triangle will be the base 1 cm/s times the height 4/3 cm/s divided by 2; multiply by the amount of time passed.

  14. anonymous
    • 4 years ago
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    and to get 2), you figure out how big the total triangle is, and then divide by 2 divide by 1 cm/s, i think.

  15. Curry
    • 4 years ago
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    hey here are some flaws with our thinking

  16. anonymous
    • 4 years ago
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    ok

  17. Curry
    • 4 years ago
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    for number 4 that is true only as long as u reach the altitude after that each one cm is going to create a trapezoid

  18. Curry
    • 4 years ago
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    same with number 2

  19. anonymous
    • 4 years ago
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    good point

  20. anonymous
    • 4 years ago
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    that's why part 1 is important. it tells you when you need to switch over.

  21. Curry
    • 4 years ago
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    at 3.6

  22. anonymous
    • 4 years ago
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    if you're in the US, technically i think it's a trapezium, not a trapezoid; since it won't have parallel sides lol

  23. Curry
    • 4 years ago
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    im in the the us we just refer to it as trap in my class

  24. Curry
    • 4 years ago
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    since it uses the same formula i believe

  25. anonymous
    • 4 years ago
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    i learned (b1 + b2)*h/2. i am very curious as to what b2 will be. however, there's an easier way, and one that's already in place; split it into 2 triangles. the second triangle will have a new area/s equation.

  26. anonymous
    • 4 years ago
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    have you learned about piecewise functions yet?

  27. anonymous
    • 4 years ago
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    it's been a long time since i had precalc

  28. Curry
    • 4 years ago
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    ye

  29. Curry
    • 4 years ago
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    number 4 doesnt work for 2/3

  30. Curry
    • 4 years ago
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    cause the rea has to be greater than any given side

  31. Curry
    • 4 years ago
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    area*

  32. anonymous
    • 4 years ago
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    is this electronic submission? sounds like what 4 is asking for is an equation.

  33. Curry
    • 4 years ago
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    it is asking how much area is covered for every centimeter

  34. anonymous
    • 4 years ago
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    i may be misunderstanding you, but the area swept out changes with every centimeter. this is time dependent, not distance dependent... also remember that k is moving at one CENTImeter per second, and the triangle has sides in meters.

  35. Curry
    • 4 years ago
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    yes

  36. anonymous
    • 4 years ago
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    ok. didn't mean to be condescending or anything..

  37. Curry
    • 4 years ago
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    but if it moves 1 centimeter then thats 1 sec after 1 centimeter the 2 legs are 1cent imeter and 4/3 cm which creates an area of 2/3 but the area has to be greater than the sides

  38. Curry
    • 4 years ago
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    for all 3/4/5 traingles

  39. anonymous
    • 4 years ago
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    oh i see what you're saying now.

  40. anonymous
    • 4 years ago
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    that somehow doesn't seem right, though. if you take 3/4/5 and then divide all sides by two to get 1.5/2/2.5, if you multiply 1.5 by 2 and then divide by 2, you get 1.5; am i thinking about this wrong?

  41. Curry
    • 4 years ago
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    ye i see what ur saying

  42. anonymous
    • 4 years ago
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    still, i am doing something wrong.

  43. Curry
    • 4 years ago
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    see thats my problem, but is that really possible

  44. Curry
    • 4 years ago
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    wait u r doing something wrong

  45. Curry
    • 4 years ago
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    wait nvmthats the same thing

  46. anonymous
    • 4 years ago
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    ok, here's the problem: when you increase the distance along AC by 1, you're not increasing the area by 2/3cm^2, you're actually increasing the *length of k* by 2/2 cm

  47. anonymous
    • 4 years ago
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    2/3* cm

  48. Curry
    • 4 years ago
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    then the area would be even smaller

  49. anonymous
    • 4 years ago
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    huh.

  50. anonymous
    • 4 years ago
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    lol

  51. Curry
    • 4 years ago
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    1 * 2/3 /2

  52. Curry
    • 4 years ago
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    is 2/6

  53. anonymous
    • 4 years ago
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    that was a "huh, you're right" not a "huh?"

  54. Curry
    • 4 years ago
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    kk

  55. Curry
    • 4 years ago
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    such a carzy problem , my teacher solved it 10 min

  56. Curry
    • 4 years ago
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    but refused to chare this with us

  57. Curry
    • 4 years ago
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    we need sataleite

  58. Curry
    • 4 years ago
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    or zarkon

  59. anonymous
    • 4 years ago
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    zarkon?

  60. anonymous
    • 4 years ago
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    i'm dumb. dumb dumb dumb. i think it's going up by 4/3 not 2/3

  61. Curry
    • 4 years ago
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    still doesnt work

  62. anonymous
    • 4 years ago
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    hmm

  63. Curry
    • 4 years ago
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    wow no one can answer this

  64. Curry
    • 4 years ago
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    wow no one can answer this

  65. razor99
    • 4 years ago
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    so u r really happy about it,

  66. Curry
    • 4 years ago
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    ??

  67. anonymous
    • 4 years ago
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    OK, I'm commenting so that I have this saved for tomorrow. Curry, post if you don't get help while I'm asleep. I'll bring it to the higher-math helpers' attention tomorrow.

  68. Curry
    • 4 years ago
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    kk im a go to bed too.

  69. anonymous
    • 4 years ago
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    currry do you have answers?

  70. Curry
    • 4 years ago
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    im sure this is one of those prob lems that only zarkon or satelite can solve i have the answer for 1 which im sure baout for 3 i got 10.7241

  71. Curry
    • 4 years ago
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    4 and 2 are insanely unthinkable

  72. Curry
    • 4 years ago
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    ig u guys figure out the answer pleazse post

  73. Curry
    • 4 years ago
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    ig u guys figure out the answer pleazse post

  74. Curry
    • 4 years ago
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    check my answer for number 3

  75. Curry
    • 4 years ago
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    oh wait thats is true

  76. Curry
    • 4 years ago
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    so number 2 woulb be 500 secsonds

  77. Curry
    • 4 years ago
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    so number 2 woulb be 500 secsonds

  78. Curry
    • 4 years ago
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    hey fool if u happen to find all the answers plz post it with the WORK. I got these answers plz do check them 1.6min 2.8.3sec 3.10.7241 as the area 4. ?

  79. Curry
    • 4 years ago
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    2. 8.3sec

  80. Curry
    • 4 years ago
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    3. 10.7241

  81. Curry
    • 4 years ago
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    i will check in the morning for answer. its 1am for me so im going to b ed

  82. anonymous
    • 4 years ago
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    as far as i can tell, 6 minutes is correct.

  83. anonymous
    • 4 years ago
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    8.3 seconds can't be right, though.

  84. anonymous
    • 4 years ago
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    yeah, but that's not question 1 haha. question 1 is just asking how long it takes for k to equal the height. *edited for spelling/grammar*

  85. anonymous
    • 4 years ago
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    taking the hypotenuse as the base.

  86. anonymous
    • 4 years ago
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    |dw:1328083228792:dw|

  87. anonymous
    • 4 years ago
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    ok, now let's get back to part 2. first, we have to note what we're looking for; we want the time when the area swept out by k is half of the total area. the total area is 6*8/2 = 24, so we want 24/2 = 12. easy part done. we know just by looking that it's going to be past point B. so we can include that area as already accounted for, and just look for the remainder.

  88. anonymous
    • 4 years ago
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    hopefully this isn't too confusing, but we have area 1 as follows: |dw:1328085095628:dw| and we want to figure out how big area 2 needs to be: |dw:1328085234534:dw|

  89. anonymous
    • 4 years ago
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    As you already figured out, Curry, and I obtusely (oh, haha! a math joke...sorry i'll stop now) didn't understand, we have a trapezoid as the second area. Area 1 is, of course, 4.8*3.6/2 = 8.64 m^2 Area 2, therefore, needs to be 12 - Area1 = 3.36 m^2 we'll set this equal to the equation for the trapezoid, which is (k+B2)*x/2 = 3.36

  90. anonymous
    • 4 years ago
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    lol u r still here

  91. anonymous
    • 4 years ago
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    we know that k = 4.8 m, so we need to know what B2 is so that we can solve for x. |dw:1328085662851:dw| this is a little harder than figuring out the growth of the triangle segment.

  92. anonymous
    • 4 years ago
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    one way to think about it is to take the imaginary little triangle at the top and figure out it's growth instead |dw:1328085960597:dw| for each increment of 1 cm to the right, the small triangle will grow 3/4 cm in height. since we want B2, we'll take \[k - 3/4 \times x\]

  93. anonymous
    • 4 years ago
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    The area of the trapezoid will therefore be (k + (k-(3/4)*x)/2 i don't like their equation builder so i'm going to stop using it until i learn it better.

  94. anonymous
    • 4 years ago
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    set that equal to 3.36 m, and we have a quadratic: .375x^2 - 4.8 x + 3.36 = 0. solve this for x to get 12.06 m and .743 m we can throw out 12.06 because that would put us beyond the end of the original triangle, but .743 is viable. finally, to get the answer for 2, we add .743 to 3.6, and multiply by 100 to get the number of seconds: 434.3 seconds., or 7 minutes 14ish seconds.

  95. anonymous
    • 4 years ago
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    4.8 meters.

  96. anonymous
    • 4 years ago
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    so 3.6/4.8/6 triple

  97. anonymous
    • 4 years ago
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    i THINK that's 1 and 2 conclusively solved now. @curry sorry for long-winded answer. 3 and 4 seem poorly worded.

  98. anonymous
    • 4 years ago
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    if 3 is asking: "what is the total area swept out by k after 6:45", the answer is Area1 = 8.64, Area2 = 2.084 Area1 + Area2 = 10.7241 meters. so looks like he got that part.

  99. anonymous
    • 4 years ago
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    So now, we have to ask again what 4) is really looking for... cuz the way i read it, sounds like it wants an equation. the area swept out per centimeter is constantly changing.

  100. anonymous
    • 4 years ago
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    If it wants an equation, it's a bit challenging but doable: By sneaky algebraic methods, the area can be described as: Area = (2/3)*x^2 for x <3.6 Area = 8.64 + 4.8*(x-3.6)+.375*(x-3.6)^2 for x>3.6 The -3.6 in there should shift the equation correctly so that everything is accounted correctly. I think there's an even sneakier way to combine the two equations into one, but i honestly can't remember/figure out how to do it. it's after 4 am where i live. I encourage you to plug in various values for x between 0 and 10 to see if this works correctly! If it wants a flat number, I'm afraid I don't know how to help you... I hope that the rest of this proves useful, and good luck with your assignment!

  101. anonymous
    • 4 years ago
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    for reference: x = 0 should give 0, x = 3.6 should give 8.64, x = 4.343 should give about 12, and x = 10 should give 24.

  102. dumbcow
    • 4 years ago
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    just to add to the conversation: I found an equation for area thats a bit less complicated x = t/100, where t is in sec |dw:1328091690260:dw| From the 2 similar triangles we can make the following relationship \[\frac{10-x}{y} = \frac{8}{10} \rightarrow y = \frac{5(10-x)}{4}\] The 3rd side of the triangle then would be \[\frac{3(10-x)}{4}\] Area of the triangle \[\rightarrow \frac{1}{2}(10-x)*\frac{3(10-x)}{4} = \frac{3}{8}(10-x)^{2}\] Now total area of large triangle is 24, so area of part the line has passed is the difference for 0<x<3.6 \[A = \frac{2}{3}x^{2}\] for 3.6 < x <10 \[A =24 - \frac{3}{8}(10-x)^{2}\]

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