Let
![y=x^{-1}](https://tex.z-dn.net/?f=y%3Dx%5E%7B-1%7D)
, so that
![y^2=x^{-2}](https://tex.z-dn.net/?f=y%5E2%3Dx%5E%7B-2%7D)
. Then
![6x^{-2}-7x^{-1}+1=0\iff6y^2-7y+1=0\implies (6y-1)(y-1)=0](https://tex.z-dn.net/?f=6x%5E%7B-2%7D-7x%5E%7B-1%7D%2B1%3D0%5Ciff6y%5E2-7y%2B1%3D0%5Cimplies%20%286y-1%29%28y-1%29%3D0)
which has two solutions,
![y=\dfrac16](https://tex.z-dn.net/?f=y%3D%5Cdfrac16)
and
![y=1](https://tex.z-dn.net/?f=y%3D1)
. So the solutions in terms of
![x](https://tex.z-dn.net/?f=x)
would be
![x^{-1}=\dfrac16\implies x=6](https://tex.z-dn.net/?f=x%5E%7B-1%7D%3D%5Cdfrac16%5Cimplies%20x%3D6)
It is the first one on the left ( 4 divided by 7)
Given:
(3,6) and (4,10)
a) f(x) = 4x - 6
f(3) = 4(3) - 6 f(4) = 4(4) - 6
f(3) = 12 - 6 f(4) = 16 - 6
f(3) = 6 f(4) = 10
b) f(x) = x + 4
f(3) = 3 + 4 f(4) = 4 + 4
f(3) = 7 f(4) = 8
c) y = x + 4
y = 3 + 4 y = 4 + 4
y = 7 y = 8
d) y = 4x - 6
y = 4(3) - 6 y = 4(4) - 6
y = 12 - 6 y = 16 - 6
y = 6 y = 10
Among the choices, choice A and D have the same value. However, the problem says functions notation. Choice A. is the function notation. A.) f(x) = 4x - 6
C. 2
An integer is a full number that is not a fraction :)
Answer:
it's a. ctxr6irx5ixtxtcycyd TY ibis4dyhugbyrobut