Write the expression as a single logarithm log3 40-log3 10
2 answers:
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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:
Let x be the number of 2 coins and y be the number of 5 coins. Let z be the number of 10 coins she exchanged. Then, we have the first equation
Since she has one 5 coin less than 2 coins, then y=x-1. So we get the equation
which is equivalent to
If we add 5 on both sides and factor out on the right by 5 we get
We can solve for x and get
Since x represents the number of 2 coins, we must have that x is an integer. That is, we cannot have, for example 1.5 2 coins. Since 7 doesn't divide 5, we must have that 7 divides 2z+1. That is, the value of z is such that the number 2z+1 is multiple of 7. That is 2z+1 = 7p where p is an integer.
If that is the case, then
x = 5p, then y = 5p-1. By having 2z+1 = 7p, we are forcing that p is an odd integer, so we have infinite solutions. For each value of p that is an odd integer, then x=5p and y=5p-1 is a solution to the problem