x > 100;

x < 150;

(x-1)/2 + 1/2 = x/2;

(x-1)/2 + 1/2 = x/2;

(x-1)/3 + 1/3 = x/3;

(x-1)/4 + 1/4 = x/4;

(x-1)/5 + 1/5 = x/5;

(x-1)/6 + 1/6 = x/6; 

 The answer is 121, why doesn't it solve it?

Solve what?

I'll look at the last one.

(x-1)/6 + 1/6 = x/6;       Multiply both sides by 6

x-1+1=x

x=x

Well that is true fro all values of x.

The others all work out the same.                       

so the answer must be  B 

right ? 

Ive given you points saseflower but I think a lot of people would not understand what you mean.

Maybe you could have written it like this

if (x1,y1) and (x2,y2) are 2 points on the graph then

the slope (gradient) of the line is (y2-y1)/(x2-x1)

There are errors in here.  Can you find them?       

This is a great answer thank you.  

The period of a wave is the time taken to complete one cycle.

You may find this clip interesting.

There is no hightest number.  You can always add more.  

x-5y=3

-5y=-x+3

y=(-1/-5)x + (3/-5)

y=(1/5)x    -3/5

gradient is 1/5   

yint = -3/5

I think standard form is the English way of saying scientific notation.

so I think the answer that is wanted is     $$1.5\times 10^{7}$$

2 you dumbo

First I will state that I am doing this by 'ear' my technique may not be as 'neat' as others may provide.

I think my answer is correct.  :)

$$\\\sqrt3\div 8-1\div8i\\\\
\frac{\sqrt3}{8}-\frac{1}{8}i$$

$$\\rcos\theta=\frac{\sqrt3}{8}\qquad rsin\theta=-\frac{1}{8}\qquad r>0\\\\
\frac{rsin\theta}{rcos\theta}=\frac{-1}{8}\div\frac{\sqrt3}{8}\\\\
tan\theta=\frac{-1}{\sqrt3}\\\\
Since\; cos\theta $ is pos and $sin\theta$ is neg $\theta $ must be in the 4th quadrant $\\\\
\theta=\frac{-\pi}{6}\\\\
$Now find r$\\\\
rcos\theta=\frac{\sqrt3}{8}=r\times\frac{\sqrt3}{2}\\\\
\frac{\sqrt3}{2}*\frac{1}{4}=r\times\frac{\sqrt3}{2}\\\\
r=\frac{1}{4}=0.25\\\\
so\\\\$$

$$\\\frac{\sqrt3}{8}-\frac{1}{8}i=0.25(cos\frac{11\pi}{6}+isin\frac{11\pi}{6})$$

anon has converted a mixed numeral into an improper fraction.  :)

Thanks MamaLlama  

$$\\2^{21014} -2^{213}\\
=2^{20801+213} -2^{213}\\
=2^{213}(2^{20801} -1)\\$$

I am just playing with this the numbers are too big to be calculated using normal means.

$$\\\displaystyle\lim_{t\rightarrow 0}\: \;\frac{tan12t}{sin4t}\\\\\\
=\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin12t}{cos12t\;sin4t}\\\\\\
=\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin4tcos8t+cos4tsin8t}{cos12t\;sin4t}\\\\\\
=\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin4tcos8t+cos4t*2sin4tcos4t}{cos12t\;sin4t}\\\\\\
=\displaystyle\lim_{t\rightarrow 0}\: \;\frac{cos8t+cos4t*2cos4t}{cos12t}\\\\\\
=\frac{cos0+cos0*2cos0}{cos0}\\\\\\
=\frac{1+1*2*1}{1}\\\\
=3$$

$${\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{127}}$$ this goes to numerator.

Your answer is $${\frac{{\mathtt{127}}}{{\mathtt{5}}}}$$

I get it now! Thanks! 

-w+164=22

-22+164=w

w=142

1. Remember the form y=mx+b.
2. Bring your function to this form.

x-5y=3

$${\mathtt{y}} = {\frac{\left({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}{{\mathtt{5}}}}$$

$${\mathtt{y}} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{x}}}{{\mathtt{5}}}}$$

$${\mathtt{y}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{5}}}}$$

Now you can see m= $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}}$$

1. Instead of comparison signs (<, <= etc) plug only =. 
2. Find y for x  =0. ( This is your y intercept.) Answer is (0,-1)
3. Find x for y = 0. ( This is your x intercept.)  Answer is (-4,0)

Hope it helps. :)

Example: 4! = 4 × 3 × 2 × 1 = 24

5.5677643628300219

what is a factorial?

It is used for factorials.