At what value of x does graph of the following function f(x) have a vertical asymptote? F(x)= 1/x+3
1 answer:
Answer:
A
Step-by-step explanation:
Given
f(x) =
The denominator of f(x) cannot be zero as this would make f(x) undefined.
Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.
x + 3 = 0 → x = - 3 is the vertical asymptote
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Answer:
Step-by-step explanation:
Given equation can be re written as
............(i)
Now it is given that y(π/2) = 2e
Applying value in (i) we get
ln(2e) = sin(π/2) + c
=> ln(2) + ln(e) = 1+c
=> ln(2) + 1 = 1 + c
=> c = ln(2)
Thus equation (i) becomes
ln(y) = sin(x) + ln(2)
ln(y) - ln(2) = sin(x)
ln(y/2) = sin(x)
Answer:
15 seconds
Step-by-step explanation:
If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:
d = rt
d+200 = 150t
And the squirrel's:
d = 100t
If we solve both for t and set them equal to each other we have:
Now we can cross multiply to solve for d:
150d = 100d + 20,000 and
50d = 20,000
d = 400
But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:
That's 1/4 of a minute which is 15 seconds.