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LynFran

  • one year ago

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  1. LynFran
    • one year ago
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  2. DecentNabeel
    • one year ago
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    \[\mathrm{Find\:the\:equivalent\:expressions\:\to\:}\left|2x-3\right|\mathrm{\:at\:}-1\le \:x\le \:2\mathrm{\:without\:the\:absolutes}\] \[-1\le \:x\le \frac{3}{2}:\quad \left(3-2x\right)\] \[\frac{3}{2}\le \:x\le \:2:\quad \left(2x-3\right)\] \[=\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx+\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx\] \[\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx=\frac{25}{4}\] \[\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx=\frac{1}{4}\] \[=\frac{25}{4}+\frac{1}{4}\] \[=\frac{13}{2}\] Answer is.. \[\frac{13}{2}\quad \left(\mathrm{Decimal:\quad }\:6.5\right)\]

  3. DecentNabeel
    • one year ago
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    any confusion @LynFran

  4. LynFran
    • one year ago
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    ok i know the answer is 13/2...but i have to use a method from geometry ...? im confuse...

  5. LynFran
    • one year ago
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    @Hero

  6. LynFran
    • one year ago
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    @campbell_st

  7. LynFran
    • one year ago
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    @TheSmartOne @abb0t @sleepyhead314

  8. LynFran
    • one year ago
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    @UsukiDoll

  9. UsukiDoll
    • one year ago
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    I don't know the geometry method.. I would've done the same thing as @DecentNabeel

  10. sleepyhead314
    • one year ago
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    |2x-3| is an absolute value function which when you graph looks like a V you would graph it then us the triangle formula

  11. sleepyhead314
    • one year ago
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    1/2 base times height

  12. UsukiDoll
    • one year ago
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    yes absolute value graphs have a V shape. ^ :)

  13. sleepyhead314
    • one year ago
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    would have to break into two triangles

  14. LynFran
    • one year ago
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    so how do i graph it...do i need to input the lower and upper limits to get the function..im still confuse..

  15. sleepyhead314
    • one year ago
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    |dw:1436662192706:dw|

  16. sleepyhead314
    • one year ago
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    |dw:1436662409249:dw|

  17. sleepyhead314
    • one year ago
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    we can get two triangles one with a base of (1.5 + 1) and height of 5 another with a base of (2 - 1.5) and height of 1

  18. sleepyhead314
    • one year ago
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    sorry that my drawing is not really scaled ^_^"

  19. LynFran
    • one year ago
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    where did u get those points from? |dw:1436669704611:dw|

  20. sleepyhead314
    • one year ago
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    by graphing... y = |2x - 3| y intercept is "-3" coming out of absolute value +3 x intercept is 1.5 find the points given by bounds when x = -1 --> y = 5 when x = 2 --> y = 1

  21. sleepyhead314
    • one year ago
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    you have learned how to graph absolute value functions before... right? o-o (i mean, if you're learning calculus, I'm assuming you know how)

  22. LynFran
    • one year ago
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    so i dont need to input the upper and lower limits in |2x-3|...thats what im not clear no...

  23. sleepyhead314
    • one year ago
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    naaa you just need to be able to visualize it |dw:1436662834072:dw|here are the bounds

  24. sleepyhead314
    • one year ago
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    here are the triangles you need to find the geometric areas of |dw:1436662904906:dw|

  25. LynFran
    • one year ago
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    ok thanks

  26. sleepyhead314
    • one year ago
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    need me to go further or you got it?

  27. LynFran
    • one year ago
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    i got it thanks

  28. sleepyhead314
    • one year ago
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    back to sleep then have fun zzzzzzzzzzzzzzzzzzzz

  29. LynFran
    • one year ago
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    no wait..look at this one how they get those coordinate for that sketch ?..didnt they input the limit to get it? https://www.symbolab.com/solver/calculus-calculator/%5Cint_%7B-1%7D%5E%7B4%7Dx%20dx%20/?origin=enterkey @sleepyhead314

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