Answer:
Below.
Step-by-step explanation:
f(x)=3x^2-6x-45/x^2-5x
= 3x^2-6x-45 / (x(x - 5))
The denominator is zero when x - 0 and x = 5.
So there will be a vertical asymptote at x = 0 and a hole at x = 5.
If we do the division we get f(x) = 3 remainder 9x - 45
so there will also be a horizontal asymptote where x = 3.
The question is incomplete, here is the complete question:
Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.
<u>Answer:</u> The half life of the sample of phosphorus-32 is 
<u>Step-by-step explanation:</u>
The equation used to calculate the half life of the sample is given as:

where,
m(t) = amount of sample after time 't' = 356 g
= initial amount of the sample = 500 g
t = time period = 7 days
h = half life of the sample = ?
Putting values in above equation, we get:

Hence, the half life of the sample of phosphorus-32 is 
Answer:
15 minutes
There are 20 quarter hours in 5 hours. This is solved by first breaking down an hour into quarters. A quarter is 1/4 of an hour, or 15 minutes.
Answer:
d..
Step-by-step explanation: