basic maths

5/6-0
____
3⅓-0

=m but how to get m plz help

are you saying

m = (y^2 - y) / (x^2 - x)

? And they're asking you to plug in values for x and y?

>>8382413
He's literally asking how to solve the slope formula.
This is obviously a troll.

>>8382413
yes the formula is m = (y^2 - y) / (x^2 - x)
and the values of x and y are:
x1=0
y1=0
x2=3⅓
y2=5/6

How to get to answer that is 1/4 , i know the answer i just dont know how to get it from the formula... :'( if some1 could explain in detail how to do it thatd be great !

What about finding a side in a triangle with only one angle and a side known?

>>8382455
lol i got 100% in this but i cant fucken find the m value ...

>>8382472
Tell your teacher to raise her standards.

>>8382410
y=mx

This is a straight line. Any value of y will equal a single value of x.

Pick any two points on the line.

Find the difference between their heights on the y axis. Find the difference between the distances on the x axis.
Divide the difference in y by the difference in x and you get m.

That's because m is the rate in change of y with respect to x.

You can call it y per x like you would call speed miles per hour (the change in distance, miles, with respect to time.)

>>8382488
I recommend you test this yourself. Draw a line on a graph and choose two points, and find m. Choose any other two points and find m again.

The m's will be the same.

This is because m describes the relationship of any x to any y.

If x increased by 1, y increases by m.

If x increases by 2, y increases by m*2

If x increases by x, y increase by m*c

>>8382472
I'll suck your balls, please tell me how.

Consider the equation of the line [math]y= mx + b[/math] in [math]\mathbb{R}^2[/math]

By picking two points in the line, [math]y_1 = m x_1+b[/math] and [math]y_2 = m x_2 +b[/math], we can solve one of the equations for b and substitute it into the other, for example, solve the second for b and substitute into the first.

[math]b = y_2 - m x_2 \Rightarrow y_1 = m x_1 + y_2 - m x_2 \Rightarrow m x_2 - m x_1 = y_2 - y_1[/math]
[math]\displaystyle \Rightarrow m(x_2 - x_1) = y_2 - y_1 \Rightarrow m = \frac{y_2 - y_1}{x_2 - x_1}[/math]

We can see if the slope is negative of positive by comparing [math]y_2,y_1,x_2,x_1[/math] given.
Since [math]y_2 > y_1[/math] and [math]x_2 > x_1[/math] it follows that [math]\displaystyle \frac{y_2 - y_1}{x_2 - x_1} > 0 \Rightarrow m > 0[/math]

We also know that [math]\displaystyle |m| = \frac{|y_2 - y_1|}{|x_2 - x_1|}[/math].

Where [math]|y_2 - y_1| = d(y_2,y_1)[/math] the distance from [math]y_2[/math] to [math]y_1[/math] and [math]|x_2 - x_1| = d(x_2,x_1)[/math], the distance from [math]x_2[/math] to [math]x_1[/math], and we also know that [math]m>0[/math], so [math]|m| = m[/math].

So we can write the our equation as [math]\displaystyle m = \frac{d(y_2,y_1)}{d(x_2,x_1)}[/math]

Now, to simplify things a bit, we take [math]\tau = \mathbb{P}(\mathbb{R}^2)[/math], so [math]\displaystyle m = \frac{d(y_2,y_1)}{d(x_2,x_1)} = \frac{d(5/6,\,0)}{d(3⅓,0)} = \frac{1}{1} = 1[/math]

So [math] m = 1 [/math]

answer is 1/4 try again plz dont fucken know how to get 1/4 from d(y2,y1)d(x2,x1)=d(5/6,0)d(3⅓,0)

found it !!!!

(5/6,0)
_____
(3⅓,0)

(3⅓,0) = 3/1 + 1/3 = 9/3 + 1/3 = 10/3


(5/6,0)
_____
(10/3,0) = 1/4

thx for youre help guys

>>8382639
That simply comes from our simplification of [math]\tau = \mathbb{P}(\mathbb{R}^2)[/math]

So d becomes [math]d(a,b) = 0[/math] if [math] a = b[/math] and [math]d(a,b) = 1[/math] if [math]a \neq b[/math]

File: HP.png (32KB, 342x401px) Image search: [Google]

32KB, 342x401px

>>8382579
I'm looking for HP and I don't know how to do it.

>>8382405
The m formula is to calculate the slope of a linear function.
Just plug in the values and do it:

3 and 1/3 is 10/3

m = 5/6-0 / 10/3 - 0
= (5/6)/(10/3)
= 5*3/6*10
=15/60
=1/4

Not that hard OP

>>8382690
Just apply as many side-angle rules as you can until the answer shakes out.

>>8382690

Find the length of HS

JH moves through a distance SH in 49 degrees
JH must then move through a distance SP in 59 degrees
Set up a ratio to find the length of PS then subtract HS to find HP

>>8382851

the answer is 26.0

>>8382690
Use the law of cosines to compute the line segment HS, then the law of sines to compute the angle HSJ. Now that you have HSJ and PJS, you know the angle on SPJ. Use the law of sines again to compute the line segment PS, and take the line segment HS away to compute PH.

File: I'm so dumb.png (12KB, 322x95px) Image search: [Google]

12KB, 322x95px

>>8382808
>>8382851
>>8382876
Thanks you guys. Holy shit what a blank my brain pulled. And it's probably not over yet. I ended up using tan like so but your replies sounds so much more right.
I'm sorry for SPAMing you I was running around in circles like a panicked chicken with his head up his ass. Why all these calculations? Was that goniometer from third grade just a waste of money I ask?

>>8382751
sorry but i just started to do maths again, after 10 years, im a little bit rusty lol and like the topic says, its basic maths no need to hate fucken faggot LOL