CPhill

(7/4) / 6  =

(7 / 4)  / ( 6  / 1)  =

(7 / 4)  * ( 1 / 6)   =

7 / 24    

or...if this is :

7 / (4 / 6)    we have

(7 /1)  / (4 /6)  =

(7 /1) * (6 / 4)  =

42 / 4   =

21 / 2

We know the hypotenuse [9] and the adjacent side [1.5]

So.....we can use the cosine inverse to solve this:

cos^-1 [ 1.5 /  9 ]   =  about 80.4°

(2 / 5)   / 3  =

(2 /5)  / (3 /1)  =

(2 / 5)  * (1 / 3)  =

2 / 15

sin(45°) = cos(45°)   =  sqrt(2) / 2

tan(45°)  = sin(45°) / cos(45°)   =   [sqrt(2)/2] / [sqrt(2)/2]    =  1

                                                  24

                                              /         \

                                             8          3

                                           /   \            \

                                         4     2            3

                                        /  \      \               \

                                      2   2      2               3

3ab+5b-6 when a=-1 and b=3

3(-1)(3) + 5(3) - 6  =

-9 + 15 - 6  =

-9  + 9   =

0

sin (y) = 0.8313

So

sin^-1(0.8313)   = y =  about 56.2°

We are looking for the cosine of an angle  whose sin = (24/25)

So.....sine  = y/r      and x =  sqrt[25^2 - 24^2]   = sqrt[49]  = 7

So  .......cos(theta)  = x/ r  =  7/25

y = -2x + 2  (1)

3x + 2y = 1  (2)

Substitution ......put (1)  into (2)

3x + 2 [ -2x + 2]  = 1

3x - 4x + 4  = 1

-x  = -3

x = 3      and    y = -2(3) + 2   = -4

Subtraction :

y = -2x + 2    →   2x + y = 2   (1)

3x + 2y = 1    (2)

Multiply (1) by 2

 4x +2y  = 4

 3x + 2y  = 1       subtract the second from the first

x  = 3         and   y  = -2(3) + 2  = -4

Graph :  https://www.desmos.com/calculator/xxjyr6ljsh

2x^2 – 6x + 3

Using the Quad Formula where   a = 2, b = -6 and c = 3, we have

[ -(-6)  plus/minus  sqrt [ (-6)^2 - 4(2)(3]  / [ 2(2)]  =

[ 6  plus/minus sqrt [ 36 - 24] / [2*2] =

[6 plus/minus sqrt (12)] / [2*2]  =

[3 plus/minus sqrt(3)] / [2]  =

[3 + sqrt(3)] / 2   →   (B)