anonymous
  • anonymous
.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TheRealMeeeee
  • TheRealMeeeee
i will help
TheRealMeeeee
  • TheRealMeeeee
^2-4y^2=64 (1) x^2+y^2=36 (2) (1)-(2) x^2-4y^2-( x^2+y^2)=36 x^2-x^2-4y^2-y^2=36 -5y^2=36 That can't be because of a^2 cant be - value (a is any number + or -) so y=0 x^2=36 so x can get +6 or -6
TheRealMeeeee
  • TheRealMeeeee
this help

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TheRealMeeeee
  • TheRealMeeeee
wen im done click Best Response
TheRealMeeeee
  • TheRealMeeeee
its A
TheRealMeeeee
  • TheRealMeeeee
wait wats your choices
TheRealMeeeee
  • TheRealMeeeee
4
TheRealMeeeee
  • TheRealMeeeee
thats A
TheRealMeeeee
  • TheRealMeeeee
4
TheRealMeeeee
  • TheRealMeeeee
its 4
TheRealMeeeee
  • TheRealMeeeee
yeah first click Best Response
TheRealMeeeee
  • TheRealMeeeee
ok wats the question
TheRealMeeeee
  • TheRealMeeeee
Since each is represented by a quadratic equation in x and y the maximum number of intersecting points is 4. Parabola: y = ax^2 + bx + c Circle: ax^2 + by^2 + c = 0 [ possibly with plain x and y terms added in ] Hyperbola: xy = c [ and variations ]
ranga
  • ranga
Both x^2 and y^2 terms are same signs (positive) and the coefficients are different in value (+2 and +4). It is an ellipse.
TheRealMeeeee
  • TheRealMeeeee
i dont know this question omg srry
ranga
  • ranga
x^2, y^2 coefficients: same sign, same value: circle : same sign, different values: ellipse : opposite sign, hyperbola only one term is square but not both (that is y^2 and x or x^2 and y) means parabola.
ranga
  • ranga
x^2 + y^2 = 25 (1) x - y^2 = -5 (2) add them up to eliminate y^2 x^2 + x = 20 x^2 + x - 20 = 0 x^2 + 5x - 4x - 20 = 0 x(x+5) - 4(x + 5) = 0 (x-4)(x+5) = 0 x = 4 or -5 x = +4 in the first and fourth quadrants x = -5 in the second and third quadrant. They want quadrant IV. so x = +4 Put x = +4 in (2) 4 - y^2 = -5 4 + 5 = y^2 9 = y^2 y = -3 or + 3 y = -3 is in quadrant IV So the solution is (4, -3) which is in quadrant IV.
ranga
  • ranga
you are welcome.
anonymous
  • anonymous
thnx

Looking for something else?

Not the answer you are looking for? Search for more explanations.