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Wow. Thanks for finding my long lost cousins XD

Thanks guys!

The answer to life the universe and everything = 42

Express the equation \({2x^2}+4x-1\)

In the form \(a{(x+b)^2}+c\)

Where \(\{a,b,c\}\in\mathbb{Z}\)

\(\begin{array}{|rcll|} \hline && 2x^2+4x-1 \\ &=& 2\cdot (x^2+2x) - 1 \\ &=& 2\cdot \Big( (x+1)^2 -1 \Big) - 1 \\ &=& 2\cdot (x+1)^2 -2 - 1 \\ &=& 2\cdot (x+1)^2 -3 \\ \hline \end{array} \)

Sorry, I mistook * for x.

t = (2 s)/(u + v) and u + v must not =0

u = (2 s)/t - v and t must not =0

v = (2 s)/t - u and t must not=0

t = (2 s)/(x (u + v)) and x (u + v) must not = 0

u = (2 s)/(t x) - v and t must not = 0

v = (2 s)/(t x) - u and t x must not = 0

You should change your name to Dumbo. You’ve asked this question twice and you are either too stupid or lazy to search for Melody’s answer. 

what is the sum of entries in row 10 of Pascal trigangle

\(\begin{array}{|rcll|} \hline (1+x)^{10} &=& \binom{10}{0} + \binom{10}{1}x \\ &+& \binom{10}{2}x^2 + \binom{10}{3}x^3 \\ &+& \binom{10}{4}x^4 + \binom{10}{5}x^5 \\ &+& \binom{10}{6}x^6 + \binom{10}{7}x^7 \\ &+& \binom{10}{8}x^8 + \binom{10}{9}x^9 \\ &+& \binom{10}{10}x^{10} \\ \hline \end{array} \)

Set x = 1:

\(\begin{array}{|rcll|} \hline (1+1)^{10} &=& \binom{10}{0} + \binom{10}{1}\cdot 1 \\ &+& \binom{10}{2}\cdot 1^2 + \binom{10}{3}\cdot 1^3 \\ &+& \binom{10}{4}\cdot 1^4 + \binom{10}{5}\cdot 1^5 \\ &+& \binom{10}{6}\cdot 1^6 + \binom{10}{7}\cdot 1^7 \\ &+& \binom{10}{8}\cdot 1^8 + \binom{10}{9}\cdot 1^9 \\ &+& \binom{10}{10}\cdot 1^{10} \\\\ 2^{10} &=& \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} \\ &+& \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \\ \hline \end{array} \)

The sum of entries in row 10 of Pascal trigangle is 2¹⁰

Given that k  is a positive integer less than 6,

how many values can k take on such that

\(3x \equiv k \pmod{6}\) has no solutions in x ?

Rewrite:

\(\begin{array}{|lrcll|} \hline 3x \equiv k \pmod{6} \\ \text{rewrite...} \\ \begin{array}{|rcll|} \hline 3x-k &=& n\cdot 6 \quad &| \quad n \in \mathbf{Z} \qquad (n \text{ is a integer}) \\ 3x &=& n\cdot 6 +k \\ x &=& \frac{n\cdot 6 +k}{3} \\ x &=& 2n+\frac{k}{3} \quad &| \quad \frac{k}{3} \text{ is a integer, if } k = 0 \text{ or } k = 3 \\ & & \quad &| \quad \frac{k}{3} \text{ is not a integer, if } k = 1,\ k = 2,\ k=4,\ k= 5 \qquad 0 < k < 6 \\ \hline \end{array} \\ \hline \end{array} \)

A bigger building? I'll say !!!

No doubt......... "The Chuch of Infinite Proportions "

Let D be the distance of the last punt......and we have

[ 30 + 20 + 27 + 29 + D ] / 5  = 30

Multiply both sides by 5 and simplify

106 + D  =   150    

Subtract 106 from each side

D  =   44 yd

Yeah....you did fix it, hectictar.......check your answer for tan(2x)  by  comparing it with

sin(2x) / cos(2x)  .......

Yep....this looks to be a pretty cool little app.....it will install as an icon but you can drag it to your  taskbar .....in my brief experience, the graphing takes a few seconds to generate, but it's worth the wait!!!

Here's a generated pic of a couple of cones....the equation is  z^2  = x^2 + y^2   

11.125 is the total answer of 89/8

It is not rounded. 

Actually the tangent one is wrong.. I did it if tan x > 0 

....Okay I think I fixed it now

csc x = 9

sin x = \(\frac{1}{9}\)

*edit* If the tangent is negative, and the sin is positive, then the cosine must be negative.

cos x = \( -\sqrt{1-(\frac{1}{9})^2}=-\frac{4\sqrt5}{9}\)

tan x = \( \frac{\sin x}{\cos x}=\frac{1}{9} \div -\frac{4\sqrt5}{9} = -\frac{\sqrt5}{20} \)

So..using the double-angle formulas:

sin (2x) = 2 sin x cos x = \(2\cdot\frac{1}{9}\cdot-\frac{4\sqrt5}{9} \mathbf{=-\frac{8\sqrt5}{81}} \)

cos (2x) = 1 - 2 sin² x = \( 1-2\cdot(\frac{1}{9})^2\mathbf{=\frac{79}{81}} \)

tan (2x) = \( \frac{2\tan x}{1-\tan^2 x}=\frac{2\cdot{ \frac{-\sqrt5}{20}}}{1- (-\frac{\sqrt5}{20})^2}=\frac{-\frac{\sqrt5}{10}}{\frac{395}{400}}\mathbf{=-\frac{8\sqrt5}{79}}\)

I'll have to solve this in the way I know how.....

Convert  6 yards to inches  =  6 * 3 * 12   =  216 inches

One line drawn from the viewing point to the circle will form a tangent to the circle......and this will form one leg of a right triangle with the hypotenuse = the viewing distance + 1/2 the diameter  = 217.5 inches...and from the viewing point, the opposite side of this triangle will =  1/2 the diameter  = 1.5 inches

So.....1/2 of the viewing angle can be expressed as

sin θ  =  1.5 / 217.5

sin^-1 (1.5 / 217.5)  = .3951°

So.....the remaining angle in this triangle is 1/2 of the angular size we are looking for =

[ 180  - 90 - .3951]  = 89.6049 °

And twice this is the angular size  = 179.2098°

This makes sense......we are viewing a relatively small circle from a relatively large distance so we should be able to see almost 1/2 of the circle....!!!!! 

Did you get your units mixed up?? A church that is 250 MILES in length and 345 MILES in width??

By simple inspection, k can take only the following values:

k =1, 2, 4 and 5, so that it has no solution in x.

The admin just added this feature.....the "guests" are not "members"......it just gives us an idea of how many people are online at any time

Thanks!

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Just set both factors to 0 and solve.....so we have

c(7c – 1) = 0

c = 0      is one solution

And

7c - 1  = 0     add 1 to both sides

7c  = 1        divide  both sides by 7

c = 1/7      and that's the other solution