Question

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Answer 1

i will help

Answer 2

^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

Answer 3

this help

Answer 4

wen im done click Best Response

Answer 5

its A

Answer 6

wait wats your choices

Answer 7

4

Answer 8

thats A

Answer 9

4

Answer 10

its 4

Answer 11

yeah first click Best Response

Answer 12

ok wats the question

Answer 13

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 ]

Answer 14

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.

Answer 15

i dont know this question omg srry

Answer 16

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.

Answer 17

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.

Answer 18

you are welcome.

Answer 19

thnx