is proved
Given that,
------- (1)
First we will simplify the LHS and then compare it with RHS
------ (2)
Substitute this in eqn (2)
On simplification we get,
Cancelling the common terms (sinx + cosx)
We know secant is inverse of cosine
Thus L.H.S = R.H.S
Hence proved
Answer:
The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)
Step-by-step explanation:
A rational equation is a equation where
where both are polynomials and q(x) can't equal zero.
1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of
in our denomiator.
So right now we have
2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.
We can model -2.5 as
So we have as of right now.
Now let see if this passes throught point (6,-3).
So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.
If we have a variable r, in the numerator that will make this applicable, we would get
Plug in 6 for the x values.
So our rational equation will be
or
We can prove this by graphing
Answer:
Step-by-step explanation:
The blue line represent the full line length.
We can add the parted black segments together and set it equal to the line.