anonymous
  • anonymous
Shemur
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Preetha
  • Preetha
@Sam or @abb0t may be able to help.
abb0t
  • abb0t
What you have here is a system of equations. Notice what they have given you: \(y=-x^2\) What that means is that you can substitute \)-x^2\) for the \(y\) in the first equation. It might help if I give them numbers A.) \(x^2 - (y - 12)^2 = 144\) B.) \(y=-x^2\) What I meant by substitution was this, plug in the y function into \(A\) wherever you see a \(y\) So, \(x^2-(\sf\color{#8C00FF }{-x^2}-12)^2=144\) now, solve for \(x\) Can you finish this now?
abb0t
  • abb0t
Once you have (if you have a solution), plug the given value of \(x\) into equation \(B\) and you should get a value for \(y\)

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abb0t
  • abb0t
You need to perform the distributive operation: \((-x^2-12)(-x-12)\)
abb0t
  • abb0t
\((-x^2-12)(-x^2-12)\)*
phi
  • phi
after expanding the square you get \[ x^2 - (-x^2 - 12)^2 = 144 \\ x^2 -(x^4 +24x^2+144)= 144\\ -x^4 -23x^2 -288= 0 \\ x^4 +23x^2 +288=0 \] if you let u = x^2, this is \[ u^2 +23u + 288= 0\] but you will have to take the square root of a complex number to get numerical values

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