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kirill115 [55]
3 years ago
7

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Mathematics
1 answer:
nalin [4]3 years ago
4 0
The answer is: cosV=39/89=0,43
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Write the standard equation for the hyperbola with the following conditions: vertices: (-2, -3) and (6, -3) foci: (-4, -3) and (
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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:
\boxed{ \dfrac{(x-2)^2}{16} - \dfrac{(y+3)^2}{20} =1}





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