It is to be solved by reminder thorem
f(x)/(x-k) will have reminder f(k),
so, f(2) = 5*(2^4) + 8 *(2^3) +4* (2^2) -5(2) +67
=5*16 + 8*8 +4*4 -5*2 +67
=80 + 64 + 16 -10 +67
= 217
The answer to this question would be the last choice (this data has no outliers)
Explanation: The reason for this is that an outlier is basically any number or value that kind of stands off or is very separated from a set of data.
For example, if I had the numbers 1,2,3,2,9,5,7,5,8,4 and 47, 47 would definitely be an outlier as it's significantly greater than the rest of the data.
The data shown in your question doesn't vary a lot though, (it's contained within a range of 65 and 80- no number seems to be radically different).
Answer:
y = sin(x- pi/2)
Step-by-step explanation:
Please see the attached image to see your graphs.
By simple inspection we can see that the sine is shifted pi/2 to the right.
We know this because sines are periodic functions with a period T = 2*pi
We can also see that the maximum point of the graph (y = 1) got shifted by a fourth of the period.
And finally, since we want to shift the graph to the right, we need to substract from the argument of the sine term.
Answer:
#4 vertex form
and
vertex (h,k)
Step-by-step explanation:
vertex form
-2(x-3)^2+32
vertex
(h,k)
(3,32)
Answer: With 11 pipes, you need 31.82 minutes to fill the tank.
Step-by-step explanation:
Let's define R as the rate at which one single pipe can fill a tank.
We know that 7 of them can fill a tank in 50 minutes, then we have the equation:
7*R*50min = 1 tank
Whit this equation, we can find the value of R:
R = 1 tank/(7*50min) = (1/350) tank/min.
Now that we know the value of R, we can do the same calculation but now with 11 pipes.
Then the time needed to fill the tank, T, is such that:
11*(1/350 tank/min)*T = 1 tank
We need to isolate T.
T = 1 tank/(11*(1/350 tank/min)) = 31.82 min
With 11 pipes, you need 31.82 minutes to fill the tank.
