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Because by definition, a triangle is a polygon with 3 sides.   :)

given that 1m/s=3.6km/h, find:

\(\begin{array}{rcll} 1\ \frac{m}{s} &=& 3.6\ \frac{km}{h} \\ \end{array}\)

a  50 mm/s in km/h

\(\begin{array}{rcll} 50\ \frac{mm}{s} \cdot \frac{ 1\ m }{ 1000\ mm } \cdot \frac{3.6\ \frac{km}{h}}{1\ \frac{m}{s} } &=& \frac{ 50 \cdot 3.6 }{1000} \cdot \frac{m}{s}\cdot \frac{s}{m}\cdot \frac{km}{h}\\ &=& 0.18\ \frac{km}{h}\\ \end{array}\)

b 288 km/h in m/s

\(\begin{array}{rcll} 288\ \frac{km}{h} \cdot \frac{1\ \frac{m}{s} } {3.6\ \frac{km}{h}} &=& \frac{288}{3.6} \cdot \frac{km}{h} \cdot \frac{m}{s} \cdot \frac{h}{km} \\ &=& \frac{288}{3.6}\ \frac{m}{s} \\ &=& 80\ \frac{m}{s} \\ \end{array}\)

2/3x +14+1/3x

I just entered this into the web2.0 calculator and it gave me the answer no problems.  :)

I did have to put the * for times where it was needed.  That is all though ://

2/3*x +14+1/3*x = x+14

=gay

Beats me.

There are 20 b***s in the red bag and 21 b***s in the blue bag

 There is 41C20 combination are possible

269128937220 = 2.6912893722e11

That is a lot of combinations!

I've an account, but I'm unable to log in.

Been like that for months.

0.7X+4+0.3X

The probability of missing any of 1 to 60 in 100 is 40/100 or 0.4, so the probability of missing this 9 times in a row is 0.4⁹ = 0.000262144

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Wolfram Alpha says it's unknown, but see http://mathforum.org/library/drmath/view/51617.html

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Because log_mn =  p  means m^p = n  then log₂(5*x+4)=-1  means  5x + 4 = 2^-1

5x + 4 = 1/2 

5x = -7/4 

x = -7/20

cos(60°) = 1/2  so  cos³(60°) = 1/2³ → 1/8  (or 0.125)

11000 people each shake hands with 10999 people so it seems at first as though there would be 11000*10999 handshakes.  However, each handshake involves two people so the total number of handshakes is only 11000*10999/2

Nella travelled 1/3 of a journey by bus, 2/3 of the remainder by bicycle and the remaining 9km on foot.

j = journey

c = cycle

b = bus

(a) how far did nella cycle

\(\begin{array}{lrcll} (1) & c &=& (\frac23\cdot j) \cdot \frac23 \\ & \frac23\cdot j &=& \frac32\cdot c \\\\ (2) & c + 9\ \text{km} &=& \frac23 \cdot j \\ & c + 9\ \text{km} &=& \frac32\cdot c \\ & \frac32\cdot c - c &=& 9 \\ & \frac12\cdot c &=& 9 \\ & c &=& 18\ \text{km}\\ \end{array}\)

Nella cycle 18 km

(b) what was the total distance travel

\(\begin{array}{lrcll} & \frac23\cdot j &=& \frac32\cdot c \\ & j &=& \frac32\cdot \frac32\cdot c \\ & j &=& \frac32\cdot \frac32\cdot 18\ \text{km} \\ & j &=& 40.5\ \text{km} \end{array}\)

The total distance travel was 40.5 km

\(\begin{array}{rcll} b &=& \frac13 \cdot j \\ b &=& \frac13 \cdot 40.5\ \text{km} \\ b &=& 13.5\ \text{km} \\ \end{array}\)

Not the question by the other guest above 

Can you answer my legit question, I know you guys socialize and what not but I just need quick help. I answer other peoples questions when I can.

What is your guys problem with guests? 

It does seem like there is a lot of questions from guest.

Well there is a lot of irrelevant questions, for example... Deez nuts, whats nine plus ten, etc.

Well, mostly guests come here to ask actual questions, so there are a lot more questions from guests.

((4/3)*3.14*7)^3 = 25170.930706962962963

continuation

\(\boxed{~ \begin{array}{rcll} x_{1,2} &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] \pm \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\\\ && y_a = a\cdot x_a^2+b\cdot x_a+c \\\\ && f'(x_a) = 2a\cdot x_a +b \\\\ x_1 + x_2 &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] + \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\ && +\dfrac{ -[~ 1+b\cdot f'(x_a) ~] - \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } + \dfrac{ -[~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& \dfrac{ -2\cdot [~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& -\dfrac{ [~ 1+b\cdot f'(x_a) ~] } { a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& -\frac{1}{a} \cdot \left[ \dfrac{ 1 } { f'(x_a) } + b \right] \qquad | \qquad f'(x_a) = 2ax_a+b \\\\ x_1 + x_2 &=& -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \qquad | \qquad x_1 = x_a \qquad x_2 = x_b \\\\ x_a + x_b &=& -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ x_b &=& -x_a -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ \end{array} ~}\)

conclusion

\(\boxed{~ \begin{array}{lcll} \text{Parabola } \quad y = ax^2+bx+c \qquad \text{ and we have } \quad a,b,c \text{ and } x_a \\\\ \Rightarrow x_b = -x_a -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ \Rightarrow y_b = ax_b^2+bx_b+c \\\\ \end{array} ~}\)

Example:

\(\begin{array}{rcll} \text{Parabola } \quad y &=& \frac12 x^2 \qquad a=\frac12 \qquad b=0 \qquad c=0 \qquad x_a=4 \\\\ x_b &=& -4 -\frac{1}{\frac12} \cdot \left( \dfrac{ 1 } { 2\cdot \frac12 \cdot 4+0 } + 0 \right) \\\\ x_b &=& -4 -\frac{1}{\frac12} \cdot \left( \dfrac{ 1 } { 4 } \right) \\\\ x_b &=& -4 - \frac24 \\\\ x_b &=& -4 - \frac12 \\\\ \mathbf{x_b} & \mathbf{=} & \mathbf{-4.5} \\\\\\ y_b &=& \frac12 \cdot (-4.5)^2+0\cdot (-4.5)+0 \\\\ \mathbf{y_b} &\mathbf{=}& \mathbf{10.125} \\\\ \end{array}\)

Thanks!

Yes sir... or ma'am. 

So it would be a divergent series?