The first graph is linear and the 2nd one is nonlinear
the first table goes for linear and the 2nd table goes for nonlinear :)
Answer:
h, j2, f, g, j1, i, k, l (ell)
Step-by-step explanation:
The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).
We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).
The horizontal asymptotes are ...
- h(x): 16x/(-4x) = -4
- j1(x): 2x^2/x^2 = 2
- i(x): 3x/x = 3
- l(x): 15x/(2x) = 7.5
- g(x): x^2/x^2 = 1
- j2(x): 3x^2/-x^2 = -3
- f(x): 0x^2/(12x^2) = 0
- k(x): 5x^2/x^2 = 5
So, the ordering least-to-greatest is ...
h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)
Answer:
D) C': ( -3, -2), D': ( -4, 1), E': ( -1, -2)
Step-by-step explanation:
Formula for this equation:
(x - 5, y - 3)
C:
(2, 1)
(x - 5, y - 3)
C': ( -3, -2)
E:
(4, 1)
(x - 5, y - 3)
E': ( -1, -2)
D:
(1, 4)
(x - 5, y - 3)
D': ( -4, 1)
Answer:
Step-by-step explanation:
Given a circle centre J
Let the radius of the circle =r
LK is tangent to circle J at point K
From the diagram attached
Theorem: The angle between a tangent and a radius is 90 degrees.
By the theorem above, Triangle JLK forms a right triangle with LJ as the hypotenuse.
Using Pythagoras Theorem:
The length of the radius, 
