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a)  A line through the origin, which is  (0, 0) , and  (10, 11)  has a slope of

[11 - 0] / [10 - 0]   =   11/10

and a line parallel to this also has a slope of  11/10 .

b)  A line that passes through  (3, 8)  and  (-1, 9)  has a slope of

[9 - 8] / [-1 - 3]   =   1 / -4   =   -1/4

and the slope of a perpendicular line  =  the negative reciprocal of  -1/4  =  4

c)   A line parallel to the  x-axis  will not "rise" at all between any two points.

its slope  =  rise / run  =  0 / run  =  0

The 1st, 3rdand 5th are correct.

I'm not sure what the difference in notation means for the 2nd and 4th

                                   [maybe someone knowledgable would like to fill me in?]

but none of the side lengths will (necessarily) be equal.

More ideas from SpaceModo:

The wikibooks artical is really good too :)

Not everythink there will work here though becasue this forum would not have all the userpackages that are required.

This is something I find quite annoying and confusing about LaTex. 

There are always differences from site to site on how it will work     

I'll add your post to our latex thread :)

Hi SpaceModo,

Thanks for this,...  even in this form I may refer to it.    

It is strange that it will not display properly.

When I go into edit mode it displays the actual LaTex output but the final version does not.

It would be becasuse you have copied and pasted it and the display formatting is not quite right....

It is still strange and disappointing though.   

Thanks Chris,

Since the polynomial is symmetrical about the x axis, I guessed that the highest power of x (degreee) was even.

I could see a power of 2 would be too small but it wasn't likely to be much bigger.

I could see that if would look kind of like a parabola. With the vertex at (0,-3)

If I had been thinking clearly I could have easily surmised that the formula would be

\((y+3)=a(x+0)^n\\ y=ax^n-3 \qquad \text{Where n is even} \)

The values of a and n could be worked out easily enough the way I did it before. 

-------------------------------------

I did think there was another routine way to work out just the degree.

I thought you knew how to do tit?

Geno is the first one I think I have seen do it.. but I thought I'd seen you and Heureka do it since then....

If ZAX is equilateral and has a perimeter of 30, then each side must  = 10

And...due to symmetry, ZL is (1/2)XA

And XA  =  10...so  ZL   = 5

So triangle ZLA is a right triangle...so using the Pythagorean Theorem :

LA  =  √ [ ZA^2   - ZL^2 ]

LA  = √ [ 10^2  - 5^2 ] 

LA  =  √ [100 - 25 ] =  √ 75  =  √25 *√ 3   =  5√3

So....the area of the quadrilateral  =  XA * LA  =   10 * 5√3  =  50√3  units^2

The angles sum to 90.....so....

[3x + 3 ] + [ 5x + 7 ]   =  90

8x + 10  = 90         subtract 10 from both sides

8x  = 80               divide both sides by 8

x  = 10

So   

5(10) + 7   =  57°

3(10) + 3  =   33°

The angles formed will be supplementary....so....

[ 9x + 9 ] + [ 6x + 6 ]  = 180

15x + 15 =  180

15 [ x + 1 ]  =  180         divide both sides by 15

x + 1  =  12     subtract 1 from both sides

x  = 11

So

The larger angle is   9(11) + 9  =  108°

The smaller angle is  6(11) + 6  =  72°

Let the smaller = x

And the larger is =  3x + 52

So

x  + 3x +  52  =  180     subtract  52 from each side  and simplify

4x  =  128     divide both sides by 4

x  = 32°  ....this is the smaller angle

3(32) + 52  =  148°   ...is the larger angle

See my answer here, Manuel :

http://web2.0calc.com/questions/help-me-please_78

I guess this is  ???

[ a - b ] /  [ a + b ]    -  {2ab ]  / [ a^2 - b^2 ]

Factor the denominator of the second fraction

[ a - b ]  /  [ a + b ]  -   [ 2ab ] /  [ (a + b) ( a - b) ]

Multiply the first fraction by  [ a - b]  in the numerator and denominator

[ (a - b) (a - b)] / [ (a + b) (a - b) ]  -  [2ab ] / [ (a + b) (a - b) ]

[ (a - b) (a - b) - 2ab ] / [ (a + b) ( a - b) ]

[ a^2  - 2ab + b^2  - 2ab ] / [ (a+ b) (a - b) ]

[ a^2  - 4ab + b^2 ]  / [ a^2  - b^2 ]

True, but I don’t use it.

your welcome man 

I need help with this

😁😁😁😁 Thank you. Needed this. 👍👍👍👍

AB  is a horizontal line.....the slope of a horizontal line   = 0

Its equation  is      y  = -2

Thanks so much! 

 3(x - 4) + 2(x^2 - x + 7) - 5(x - 1)   

Constant is

3 * (- 4 )  +  2 * 7  + (-5)* (-1)  =

-12   +   14   +    5    =

7   ......just as you found ....good job....!!!

Let's first find  g (g(x)) 

This is literally putting g(x)  into the x in  g(x)....so we have.... 

2 [ 2x - 1]  -  1    =    4x - 3

So  

g (g(x))  - g (x)   =

[ 4x - 3 ] - [ 2x  - 1 ]  =

2x  - 2    =  f(x)

So

f(3)   =  2(3) - 2     =   4

Triangle  PQR   is similar to triangle ABC

So

PQ / QR   =   AB / BC

PQ / 40  =  21 / 35      multiply both sides by  40

PQ  =  (40/35) * 21  =  (8/7) * 21  = 24  =  "a"

And

PR / PQ  =  AC / AB

PR / 24  =  28 / 21   multiply both sides by 24

PR =  24 (28/21)  =  24 (4/3)  = 32  =  "b"

Right on dude!

(7+8x)-3(2+6x+2x^2)+9(1+3x+4x^2+12x^3)-9(7-x^2-8x^3+13x^4)

8x  -  18x +  27x   =

17x

So.....the coefficient is  17 

Very nice detective work, Melody!!!

I haven't seen this done before, but it appears to work!!!

Since the polynomial is symmetric about y and appears to intercept the x axis twice, I would have guessed that it was a quadratic......but yours perfectly models the data....