I do khan academy 2 but sorry the answer
Given that the hyperbola has a center at (0,0), and its vertices and foci are on y-axis. This, the equation of the hyperbola is of the form
x²/a²-y²/b²=-1 (a>0, b>0)
In the equation, vertices are (0, +/-b) .
Thus,
b=60
Foci (0,+/-√(a²+b²))
thus
√(a²+60²)=65
hence solving for a²
a²=65²-60²
a²=625
a²=25²
hence the equation is:
x²/25²-y²/60²=-1
Answer:
x=137
y=525-137
y=388
Step-by-step explanation:
Let the student tickets be x
Let the geral Admission tickets be y
x+y=525
y=525-x
4x+6y=2876 (subsitute for y)
4x+6(525-x)=2876
4x+3150-6x=2876
-2x=-274
x=137
y=525-137
y=388
Hence, about 137 childern tickets were sold and 388 gernal admission tickets were sold.
Paul.
Standard form : Ax + By = C
y = 2/3x + 1
-2/3x + y = 1....multiply by -3
2x - 3y = -3 <=== standard form
