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anonymous

  • one year ago

YAS

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  1. anonymous
    • one year ago
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    @welshfella @Hero @acxbox22 @LegendarySadist @prowrestler @Zale101 @xavierbo2 @campbell_st @vera_ewing @bubbleslove1234

  2. anonymous
    • one year ago
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    @ikram002p @IrishBoy123 @imqwerty @AliceCullen @sammixboo

  3. anonymous
    • one year ago
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    @nincompoop @nuccioreggie

  4. campbell_st
    • one year ago
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    ok, what is a simply linear equation...?

  5. anonymous
    • one year ago
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    y=5x+3

  6. campbell_st
    • one year ago
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    next, what is a simple quadratic equation...?

  7. anonymous
    • one year ago
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    16x^2+4x+2

  8. anonymous
    • one year ago
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    what do I do now?

  9. anonymous
    • one year ago
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    hello? @campbell_st

  10. campbell_st
    • one year ago
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    well I wouldn't have considered the 2 equations simple... the quadratic is positive definite.... which means it won't cut the x-axis by anyway, you have 2 equations... next step is to solve them simultaneously y = 5x + 3 y = 16x^2 + 4x + 2 so to solve them just equate them 5x + 3 = 16x^2 + 4x + 2 then solve for x. but you have chosen 2 difficult equations... I've attached a file, its a graph of your 2 equations

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  11. campbell_st
    • one year ago
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    so you may want to rethink your equations.

  12. anonymous
    • one year ago
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    i just made up 2 random equations I don't know what equations to think of

  13. campbell_st
    • one year ago
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    ok... so start with simple you linear equation is almost ok change the y-intercept to 6 so use y = 5x + 6 now a really simply quadratic is \[y = x^2\] So equating them you get \[5x + 6 = x^2\] or \[x^2 - 5x - 6 = 0\] this is a quadratic equation that can be solved by factoring... so that is what you need to do, solve for x, when you get the 2 answers, substitute the x values into either equation to find the matching y values... so you will have 2 point of intersection... if you have to graph the curves 1 use a table of values that includes both x values or just just some graphing software line https://www.desmos.com/calculator

  14. anonymous
    • one year ago
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    thank you! @campbell_st

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spraguer (Moderator)
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