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anonymous
 one year ago
c
anonymous
 one year ago
c

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wiat...let me refiigure that line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the linear equation is y=14x3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439475598238:dw here is the graph

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dang...it didnt show...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok here is the quadratic (x4)^2+(y=4)^2=4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439475897879:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes now the line is obviously your line and the circle is the quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you plot it on the graph

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i am not really good at showing my work. im only good at doing it

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0"@MeganChase went offline"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You guys were using a circle but the question asked for a quadratic (parabola). Is any shape acceptable?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. so scratch the stuff using this equation "(x4)^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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0graph y = 2x  4 and y = 2x²  4 here https://www.desmos.com/calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. 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 \[2x^2  4 = 2x  4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how to solve that for x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 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 yvalues 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So your solutions to the system are (0, 4) and (1, 2). Which you can see from the graph https://www.desmos.com/calculator/wdb2i4kk5d
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