It's the last one!
Hope this helps!
The foci of the hyperbola with equation 5y^2-4x^2=20 will be given as follows:
divide each term by 20
(5y^2)/20-(4x^2)/20=20/20
simplifying gives us:
y^2/4-x^2/5=1
This follows the standard form of the hyperbola
(y-k)²/a²-(x-h)²/b²=1
thus
a=2, b=√5 , k=0, h=0
Next we find c, the distance from the center to a focus.
√(a²+b²)
=√(2²+(√5)²)
=√(4+5)
=√9
=3
the focus of the hyperbola is found using formula:
(h.h+k)
substituting our values we get:
(0,3)
The second focus of the hyperbola can be found by subtracting c from k
(h,k-c)
substituting our values we obtain:
(0,-3)
Thus we have two foci
(0,3) and (0,-3)
Answer:
D) R’(2,3); S’(7,3); T’(5,8); U’(0,8)
Step-by-step explanation:
The last one. If you reflect it over the x-axis, it moves from quadrant IV to quadrant I. Everything in quadrant I is positive, and the last one is the only one that has all x and y values positive.
Hamid's normal weight is 80 kg.
