Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
1. a) equation of the line :
y - intercept = -7
so, it will pass through point (0, -7)
and if we plug the value of x as 5, we get
so, it will pass through point (5, -5) too
now, just plot the points (0 , -7) and (5 , -5) and join them.
2. b) equation of line is :
here, y - intercept = 5
so the line passes through point (0 , 5)
now, Plugging the value of x = 1 we get :
so, the given line passes through point (1 , 2)
plotting the points, we can get our required line.
Answer:
y = 2x - 3
Step-by-step explanation:
y - - 3 = 2(x-0)
y + 3 = 2x
y = 2x -3
Answer:
4.6 = x the side opposite C would be 6.1
Step-by-step explanation:
you can use law of sine
you know you have a right triangle so the missing angle B = 49
4/sin41=x/sin49