Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have
First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have
We don't want the denomiator be zero because we can't divide by zero.
so
So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes
So we have a horinzontal asymptofe of 2
Answer:
5 cupcakes
Step-by-step explanation:
=1x 30/ 6
=5
I hope it helps
Thank U
Answer:
a range of values such that the probability is C % that a rndomly selected data value is in that range
Step-by-step explanation:
complete question is:
Select the proper interpretation of a confidence interval for a mean at a confidence level of C % .
a range of values produced by a method such that C % of confidence intervals produced the same way contain the sample mean
a range of values such that the probability is C % that a randomly selected data value is in that range
a range of values that contains C % of the sample data in a very large number of samples of the same size
a range of values constructed using a procedure that will develop a range that contains the population mean C % of the time
a range of values such that the probability is C % that the population mean is in that range
Answer:
t = 2.52 seconds
Step-by-step explanation:
h=139-15t-16t^2
We want to know when the ball hits the ground
That would be when h=0
0 = 139-15t-16t^2
We can use the quadratic formula to find t
t = -b ± sqrt(b^2-4ac)
----------------------
2a
where a = -16 b = -15 and c = 139
t = -(-15) ± sqrt((-15)^2-4(-16)139)
----------------------
2(-16)
t = (15) ± sqrt(225+8896)
----------------------
-32
t = (15) ± sqrt(9121)
----------------------
-32
t = 15+ sqrt(9121) t = 15- sqrt(9121)
-------------------- or -------------------
-32 -32
-3.453247707 or 2.515747707
Since time cannot be negative
2.515747707
Round to the nearest hundredth
t = 2.52 seconds
Answer:
The third one ![\sqrt[3]{15}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B15%7D)
Step-by-step explanation:
