Answer:
The answer is "MS and QS".
Step-by-step explanation:
Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.
As NR and MQ bisect each other at S
⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ
In ΔMNS and ΔQNS
MN=QN (∵ MNQ is isosceles triangle)
∠NMS=∠NQS (∵ MNQ is isosceles triangle)
MS=SQ (Given)
By SAS rule, ΔMNS ≅ ΔQNS.
Hence, segments MS and SQ are therefore congruent by the definition of bisector.
The correct option is MS and QS
Answer:
The equation is ( x² / 9 ) - ( y² / 7 ) = 1
Step-by-step explanation:
Given the data in question;
hyperbola is centered at the origin, this means h and k are all equals to 0.
Vertices: (-3,0) and (3,0)
Since y-coordinates are constant, this implies it is a hyperbola with horizontal transverse axis.
h - a = -3
0 - a = -3
a = 3
Foci: (-4,0) and (4,0)
h - c = -4
0 - c = -4
c = 4
we know that, for a hyperbola
c² = a² + b²
so
⇒ ( 4 )² = ( 3 )² + b²
16 = 9 + b²
b² = 16 - 9
b² = 7
So the equation for the hyperbola will be;
⇒ ( (x-h)² / a² ) - ( (y-k)² / b² ) = 1
so we substitute
⇒ ( (x-0)² / 3² ) - ( (y-0)² / 7 ) = 1
⇒ ( x² / 3² ) - ( y² / 7 ) = 1
⇒ ( x² / 9 ) - ( y² / 7 ) = 1
Therefore, The equation is ( x² / 9 ) - ( y² / 7 ) = 1
