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:
Answer:
answer of given question is -7/2
Answer:
1-k^2
Step-by-step explanation:
tan A= sin A/cos A
(sin A)(cos A)(sin A/cos A)=(sin A)^2
(sin A) ^2 + (cos A) ^2= 1
(sin A) ^2=1-(cos A) ^2
(sin A)^2= 1 - k^2
Answer:
Suppose that logb b = x.
bx = b (Use the definition of a logarithm.)
bx = b1
x = 1 (Use the common base property.)
So, logb b = 1.
28 + (3x -2) + 28 = 6x
28+ (3x -3x -2) + 28 = 6x - 3x
28+ (-2) + 28 = 3x
28 - 2 + 28 = 3x
26 + 28 = 3x
54 = 3x
18 = x