What are the solutions of the quadratic equation? 12x^2 - 64 = -x^4
1 answer:
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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:
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:
a
the displacement of the plane is
b
The displacement of plane from Portland to Akron with layover at Houston is
c
the vector which defines this displacement is
Step-by-step explanation:
From the question we are the
The displacement from Portland, Oregon to Houston, Texas. is
The displacement from Houston, Texas to Akron, Ohio. is a
a
The motion of the plane in question a is in the opposite direction to vector
So the displacement of the plane is
b
The motion of Portland to Houston to Akron is in the direction of both vector and
so
The displacement of plane from Portland to Akron with layover at Houston is
c
From the question we are told that the plane flies over Portland to get to Anchorage, Alaska which implies that it is moving in opposite direction to the vector
Give that distance from Portland to Houston is
and the distance from Portland to Anchorage is km
Then the vector which defines this displacement is mathematically represented as