All Answers 358504 Answers

Oooh I see. Thanks :D

You learn something new everyday. 

$${\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{3}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{20}}}{{\mathtt{3}}}} = -{\mathtt{6.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$${\frac{{\mathtt{1\,000}}}{{\mathtt{3}}}} = {\mathtt{333.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

$${\frac{{\mathtt{40}}}{{\mathtt{3}}}} = {\mathtt{13.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

333-13=320

How did you come up with that answer Alan? .-.

There is no answer to the calculation

$${\frac{{\mathtt{x}}}{{\mathtt{y}}}}{\mathtt{\,\times\,}}{\mathtt{z}} = {\frac{{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{z}}}{{\mathtt{y}}}}$$

what is 1190 divisible by

$$\small{\text{
prime factorization: $ 1190 = 2^{\textcolor[rgb]{1,0,0}{1}}*5^{\textcolor[rgb]{0,1,0}{1}}*7^{\textcolor[rgb]{0,0,1}{1}}*17^{\textcolor[rgb]{0,1,1}{1}} $
}}$$

$$\small{\text{
The number of the divisors $ = (\textcolor[rgb]{1,0,0}{1} + 1)
*(\textcolor[rgb]{0,1,0}{1} + 1)
*(\textcolor[rgb]{0,0,1}{1} + 1)
*(\textcolor[rgb]{0,1,1}{1} + 1)
= 2*2*2*2 = 16$
}}$$

$$\begin{array}{|r|c|c|c|c|c|l|l|}
\hline
n & 1 & 2 & 5 & 7 & 17 & & devisor\\
\hline
1 & x & & & & & &= 1 \\
2 & & x & & & & &= 2 \\
3 & & & x & & & &= 5 \\
4 & & & & x & & &= 7 \\
5 & & & & & x & &=17 \\
6 & & x & x & & & 2*5 &= 10 \\
7 & & x & & x & & 2*7 &= 14 \\
8 & & x & & & x & 2*17 &= 34 \\
9 & & & x & x & & 5*7 &= 35 \\
10 & & & x & & x & 5*17 &= 85 \\
11 & & & & x & x & 7*17 &= 119 \\
12 & & x & x & x & & 2*5*7 &= 70 \\
13 & & x & x & & x & 2*5*17 &= 170 \\
14 & & x & & x & x & 2*7*17 &= 238 \\
15 & & & x & x & x & 5*7*17 &= 70 \\
16 & & x & x & x & x & 2*5*7*17 &= 1190 \\
\hline
\end{array}$$

One possibility is (1 + 3 + 4 - 2)*7

Another is 1*3*4*7/2

Another is (4+3)*(1+7-2)

...

$${\left({\mathtt{290}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}} = {\mathtt{84\,348.755\: \!102\: \!040\: \!816\: \!326\: \!5}}$$

$${\frac{{\mathtt{97}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{12\,500}} = {\mathtt{12\,125}}$$

Sorry Dragon,

3-(x+2)=2x-2

-(x+2)=-1*x+-1*2=-1x-2=-x-2

so

3-(x+2)=3-x-2=3-2-x=1-x

so the  problem becomes

1-x=2x-2

add x to both sides

then add 2 to both sides

the think what to do next.  :)

ctrl+a

delete

The $30 hotel puzzle.    (Thanks Landry     24/9/14 )

This one has been around for ever - that is because it is so good!

I have hidden the answer so if you want the answer from a moderator you might have to beg.   LOL

Americans are very religions and they need to tell the difference between mass and maths so they leave off the s.

That's my theory anyway!    LOL

320

..

..

..

A sprinkler system on a farm is set to spray water over a distance of 35 meters and to rotate through an angle of 140degrees. find the area of the region that can be watered

$$\\ \small{\text{
$A=\pi r^2 $ is the area of the circle.
$ A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) $ is the area of the Region can be watered.
}}
\\
\small{\text{
$ r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2$
}}\\
\small{\text{
The area of the Region can be watered is $ \textcolor[rgb]{1,0,0}{1496.62\;m^2 } $
}}$$

Something like this perhaps?

Find an equation of the line containing the given pair of points (-5,-8) and (-9,-2)

$$\small{\text{
$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1} \Rightarrow
\boxed{y=\left( \dfrac{y_2-y_1}{x_2-x_1}\right)x+\dfrac{y_1x_2-x_1y_2}{x_2-x_1}}
$
}} \\ \\
\small{\text{
$P_1 (x_1=-5, y_1=-8)
\qquad P_2 (x_2=-9, y_2=-2)
$
}}
\\ \\
\small{\text{
y=\left( \dfrac{ (-2)-(-8)}{(-9)-(-5) }\right)x
+\dfrac{(-8)(-9)-(-5)(-2)} { (-9)-(-5) }
$
}}
\\ \\
\small{\text{
$
y=- \dfrac{ 6 }{ 4 }x
+\dfrac{ 72-10 } { -4 }
$
}}
\\ \\
\small{\text{
$
y=-1.5x -15.5$
}}$$

Check: -8 = -1.5(-5) - 15.5  ok

            -2 = -1.5(-9) - 15.5  ok

That depends on what the dots mean.

If it is in the play store I think it would be perfectly safe.

the equation c2=p can have zero, one, or two solutions. what can you say about the number of solutions of the equation c3=p?

$$\small{\text{
$c^n$ = p has always n solutions ( real and imaginary solutions together )
}}$$

There should only be 21 zeros

That's 23 numbers after the point.  

Use sin^-1 or arcsine (asin on most calculators)

$${\mathtt{angle}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.083\: \!33}}\right)} \Rightarrow {\mathtt{angle}} = {\mathtt{4.780\: \!000\: \!194\: \!675^{\circ}}}$$

0.72*10⁵ is 7.2*10⁴ in scientific notation.