Rom

\(\mbox{you've got data points }(x_i, y_i) \mbox{ that you've done a regression on to obtain }\\ \tilde{y}(x) = m x + b \\ \mbox{The standard error of the regression estimate is given by }\\ S = \sqrt{\dfrac 1 N \displaystyle{\sum_{i=1}^N}\left(\tilde{y}(x_i)-y_i\right)^2}\)

there are really only 3 ways to get excluded values

a) you can't divide by 0

b) you can't take the square root of a negative number (over the real numbers)

c) you can't take the log of a non-positive number (over the real numbers)

there are a few other cases, like inverse trig functions but above are the main ones

in this case you have 35x^3 in the denominator.

So if x=0, 35x^3=0, and then you are dividing by 0 which is forbidden.

So the excluded point here is x=0

\(29 = 12n + 5 \\ 24 = 12n \\ 2 = n\)

\(A_{circle} = \pi \left( \dfrac D 2 \right)^2 = \pi (14)^2 = 196 \pi cm^2\\ \mbox{If the circle is inscribed in the box then the box will have side length }D \mbox{ with area }D^2 \\ A_{square} = D^2 = (28)^2 = 784cm^2\)

there are a whole lot of ways to approach this.

The simplest is just brute force and looking for a pattern.  Below is a list of powers of 2 and the power

(1,2), (2,4), (3, 8), (4, 16), (5, 32), (6, 64), (7, 128), (8, 256), (9, 512), (10, 1024)

and in the one's digit we see

2, 4, 8, 6, 2, 4, 8, 6, repeated indefinitely starting at a power of 1.

So if have an array of digits=(2,4, 8, 6)

We first subtract 1 from n and then divide it by 4.  We take the remainder and add 1 and use this to index into the digits array.

So in this case we find

\((2015-1) \pmod 4 +1 = 2+1 = 3 \\ digits[3] = 8 \\ \mbox{and so the final digit of }2^{2015}=8\)

a trick?  tough to say...

suppose you have |x|

if (x>=0), whatever x might be, then |x| = x

if (x<=0), then |x| = -x

that about as simple as it can be made.

0 is incorrect.

Anything divided by 0, including 0, is undefined.

\(2x+2y=20 \\ 5x - 3y=15 \\ \mbox{Using the first equation.}\\ 2x=20-2y \\ x=10-y \\ \mbox{now we plug that value of x into the second equation} \\ 5(10-y) - 3y = 15 \\ 50-5y-3y=15 \\ -8y = -35 \\ y = \dfrac {35}{8} \\ x=10-y = 10-\dfrac {35}{8}\\ x=\dfrac {45}{8}\)

www.desmos.com

Just enter the equations in at the left and the graph is drawn for you.

The point you seek is at the intersection of the two lines.

\(1.423828125 = 1 \dfrac{423828125}{1000000000}=1\dfrac{217}{512}\)