CPhill

7)  The bisected angle  in the first triangle will always set up this relationship

10 / 24  =  ?  / 36

5/12  = ? / 36        cross - multiply

36 * 5  =  12 * ?       

180   =  12 * ?      divide both sides by 12

15  =  ? 

8)  Similarly

[ 5x - 6 ]  / 42   =  28 / 49

[ 5x - 6 ] / 42  =   4 / 7        cross-multiply

7[ 5x - 6 ]  =  42 * 4

35x - 42  =  168       add 42  to  both sides

35x  =  210   divide both sides by 35

x  =  6

Second one

Scale factor  =  1/2      Dilation Center  = (8, -6)

A'  =

[(-4, 4) - (8, - 6) ] * (1/2)  + (8, - 6) =

[ (-12, 10] * (1/2) + (8, -6)  =

(-6, 5) + (8, -6)  =

(2, -1)

B' = 

[ (4, 2)  -  (8, -6) ] *(1/2) + (8, -6) =

[ (-4, 8) ] * (1/2) +  ( 8, -6)  =

(-2, 4)  + ( 8 , - 6)

(6, -2)

C'  =

[ (-6, -4) - ( 8, -6) ] * (1/2)  +  (8, -6)  =

[ (-14, 2) ] * (1/2) + (8, -6)  =

(-7, 1)  +  ( 8, - 6)

(1, -5)

Here's a pic :

Frst one   

Scale Factor  1/2      Dilation Center(-1, 1)

We need this formula

New Point '  = [ Old Point  - Dilation Center ] * Scale Factor  + Dilation Center

Remember that    (a , b)  -  ( c, d)  =  ( a - c, b - d)

And  C(a, b)  =  (Ca, Cb)

A'  = 

[ (-4, 4)  - ( -1, 1)]  * (1/2)  +  (-1, 1) =

 (-3, 3)* (1/2)  + (-1, 1)   =

(-3/2, 3/2) + (-1, 1)  =

(-5/2, 5/2)

B' =

[ (2, 4) - (-1, 1)]* (1/2)  + (-1, 1)  =

[ (3, 3) * (1/2) + ( -1, 1)  =

(3/2, 3/2) + (-1, 1) =

(1/2, 5/2)

C'  =  [ (-4, 1) - (-1, 1) ] * (1/2) + (-1, 1)  =

[ (-3, 0) ]* (1/2) + (-1, 1 )  =

(-3/2, 0)  + (-1, 1)  =

(-5/2, 1)

Here's a pic :

I assume this is :

 (x * √ (3a) *  ³√b^2)³

Writing these in exponential form is easier, Manuel.....then we can just apply the rule that (a^m)^n  = a^(m * n)

So we have

[ x * ( 3a)^(1/2) * b^(2/3) ]  ^3  =

x^3  *  (3a)^(3/2) * b^2  =

x^3 * 3a √ [3a ] * b^2

No...1/8 is correct....I was just making a joke....LOL!!!

Don't you wish that ice cream worked that way  ??

Eat  (1/3) of a bowl....you get  "wider'

Eat 3 bowls.....you get  "narrower "

If   2 - 3sqrt (2)  is a root so is its conjugate, 2 + 3sqrt(2)

And if 1 - sqrt (3)  is a root then so is  1 + sqrt(3).....so we have

[ x - (2 - 3sqrt (2) ]  [ x -  (2 + 3sqrt(2) ] [ x - ( 1 - sqrt(3) ] [ x - ( 1 + sqrt (3) ]     simplify

[ x^2 - x(2 - 3sqrt (2) - x (2 + 3sqrt (2) + 4 - 18 ]  [ x - ( 1 - sqrt(3) ] [ x - ( 1 + sqrt (3) ]    

[ x^2 - 4x - 14 ]   [  x^2 - x(1 - sqrt (3) - x (1 + sqrt (3) + 1 - 3 ]

  [ x^2 - 4x - 14 ] [ x^2 -2x - 2 ] =

x^4  - 6x^3 - 8x^2 + 36x + 28

In a quadratic function, the domain   - the values of x covered by the graph  - is always going to be  "all real numbers"

This will not be true of the range .....[  the y values covered by the graph ].

HAHAHA!!!....that's WAY easier than my clumsy approach....thanks, EP!!!

Yep......looking at the graph is a neccessity here....LOL!!!!