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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    That looks kinda crazy, I know, but it's because of the lack of accuracy and the fact I have no graph paper. That is a circle that has an equation of (x−4)2+(y+4)2=4 and a line that intersects (or touches) the circle at the point (4,-2) with a slope of 2. The line has a y intercept of -4. So the equation for the line is y=2x−4 and the point in common, aka the solution, to that system is (4,-2). The way you would do the algebra part is to show that the point (4,-2) works in both equations to make them both true. (x−4)2+(y+4)2=4 for the point (4,-2) is as follows: (4−4)2+(−2+4)2=4 (0)2+(2)2=4 4=4

  2. anonymous
    • one year ago
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    wiat...let me refiigure that line

  3. anonymous
    • one year ago
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    The equation for the line is y=14x−3 Now, let's show that the point (4,-2) makes this equation true just like it does for the circle. Subbing in -2 for y and 4 for x, we get: −2=14(4)−3 −2=1−3 −2=−2

  4. anonymous
    • one year ago
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    the linear equation is y=14x-3

  5. anonymous
    • one year ago
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    |dw:1439475598238:dw| here is the graph

  6. anonymous
    • one year ago
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    dang...it didnt show...

  7. anonymous
    • one year ago
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    ok here is the quadratic (x-4)^2+(y=4)^2=4

  8. anonymous
    • one year ago
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    |dw:1439475897879:dw|

  9. anonymous
    • one year ago
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    yes now the line is obviously your line and the circle is the quadratic

  10. anonymous
    • one year ago
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    you plot it on the graph

  11. anonymous
    • one year ago
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    sorry, i am not really good at showing my work. im only good at doing it

  12. OregonDuck
    • one year ago
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    "@MeganChase went offline"

  13. anonymous
    • one year ago
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    You guys were using a circle but the question asked for a quadratic (parabola). Is any shape acceptable?

  14. anonymous
    • one year ago
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    ok. so scratch the stuff using this equation "(x-4)^2+(y=4)^2=4" but keep the line y = 2x - 4 as the linear equation. For a quadratic equation, keep it simple and just square the x to get y = 2x² - 4

  15. anonymous
    • one year ago
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    graph y = 2x - 4 and y = 2x² - 4 here https://www.desmos.com/calculator

  16. anonymous
    • one year ago
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    it is, but just so you know the solution to that isn't rational so you're going to have to deal with square roots/decimals

  17. anonymous
    • one year ago
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    decimals*

  18. anonymous
    • one year ago
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    ok. either way the process is the same. since they're both y = something, you can use substitution and drop the y's y = 2x² - 4 and y = 2x - 4 are the equations so we can say \[2x^2 - 4 = 2x - 4\]

  19. anonymous
    • one year ago
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    Do you know how to solve that for x?

  20. anonymous
    • one year ago
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    right so now subtract 2x from both sides 2x² - 2x = 0 Factor out 2x 2x(x - 1) = 0 Set each factor equal to 0 and solve 2x = 0 → x = 0 x - 1 = 0 → x = 1

  21. anonymous
    • one year ago
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    you follow?

  22. anonymous
    • one year ago
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    yes for part 2 it looks like all you have to do is graph the line and parabola and make sure you show the intersection points because that is the solution. We still have to find the y-values for part 1 though. Just plug in 0 and 1 into either equation to get them. For x = 0: y = 2(0) - 4 For x = 1: y = 2(1) - 4 And that will give the 2 intersection points/solutions

  23. anonymous
    • one year ago
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    I don't know that it matters. They mean the same thing, just different notations. You can use y for both, or you can use function notation. If you use function notation and call the linear function f(x) then call the quadratic g(x). It doesn't matter as long as you're consistent throughout the problem

  24. anonymous
    • one year ago
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    So your solutions to the system are (0, -4) and (1, -2). Which you can see from the graph https://www.desmos.com/calculator/wdb2i4kk5d

  25. anonymous
    • one year ago
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    you're welcome

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