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:
I would say go with with B
Answer:
x = 55, y = 70, z = 125
Step-by-step explanation:
125 and x are adjacent angles and are supplementary, then
x = 180 - 125 = 55
The triangle is isosceles ( 2 congruent sides ) thus the base angles are congruent, both 55
The sum of the 3 angles in the triangle = 180° , thus
y = 180 - (55 + 55) = 180 - 110 = 70
55 and z are adjacent and supplementary, so
z = 180 - 55 = 125
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➷ 90 is an integer, and the closest others are 89 and 91
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Answer:
there is no quick way around this. U need to plug in -4,2,8 into x value into the function, and see what y returns to
Step-by-step explanation: