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x + y  =  10    square both sides

x^2 + 2xy + y^2 =  100

x^2 + y^2  =  100 - 2xy

64 + xy  = 100 - 2xy

3xy  = 36

xy  = 12

y = 12 / x

x + 12 / x  = 10

x^2 + 12  = 10x

x^2  - 10x  =  -12

x^2 -10x + 25  = -12 + 25

(x - 5)^2  =  13     take  both roots

x - 5  = sqrt (13)                            x  - 5   =  -sqrt 13

x = 5 + sqrt (13)                            x  =  5  - sqrt (13)

y  = 12 / ( 5 + sqrt 13)                   y =  12 / (5 -sqrt 13)

y = [12 (5 -sqrt 13)] / 12                y=  12 (5 + sqrt 13) / ( 12)

y = 5 -sqrt (13)                              y = 5 + sqrt 13

(x, y)   =  ( 5 + sqrt 13 , 5 -sqrt 13 )  and  ( 5 -sqrt 13 , 5 + sqrt 13)

Call his salary,  S

S - (1/5)S - (2/3)S =   (2/15)S =   amount he had after food and  transport

(2/15)S  - (1/2)(2/15)S  =  350

(2/15)S - (1/15)S = 350

(1/15)S = 350

S = 5250

Amt spent on food and  transport  = (13/15) (5250)   = $4550

y = x  slope = 1  

y = 2x   slope =  2

m  =   tan  [   [ arctan (1)  + arctan (2) ] /  2  ]  ≈  1.388

Simplify  as

x^2 + sx  +  100 

To be a square of a binomial the discriminant must  =  0

s^2   -  4(1)(100)  = 0

s^2  =   400    take both roots

s  = -20 , 20

x^2 - 4x + y^2 - 2y  = 0          complete the square on x,y

x^2 - 4x + 4 + y^2 - 2y + 1 =  4 +1

(x- 2)^2 + ( y - 1)^2  =  5

This is a  circle centered at ( 2 , 1)  with a radius of sqrt 5

The largest value of x =  2 + sqrt 5

a + b =   4          square both sides

a^2 + 2ab + b^2 =  16

a^2 + b^2   = 16   - 2ab    (1)

  And we are given that

a^2 + b^2  = 6 + 2ab        (2)

Subtract (1) from ( 2)

0 = -10 + 4ab

10 = 4ab

ab = 10 / 4 =  5/2

a^3 + b^3 =  (a + b)  ( a^2 + b^2  - ab)    = 

(4) ( 6 + 2ab - ab)  =

(4) ( 6 - ab)  =

(4) ( 6  - 5/2)  =

(4) ( 7/2)  = 

  14

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}} \)

Working  from  the bottom up

4 + 1/2 =  9/2

1/ (9/2)  = 2/9

4 + 2/9  =  38/9

1/ (38/9) =  9/ 38

1 + 9/38 = 47/38

1 / (47/38)  =  38 / 47

625^x  =  9   

(5^4)^x  =  9           take each side to the 3/4   power

(5^4)^(3/4 * x)   = 9^(3/4)

(5^3) ^x   =  9^(3/4)

125^x  =   ⁴√(9^3)

125^x  = ⁴√729

A =  2000 ( 1 + .06/12)^(4 * 12)  =  2000 (1 + .06/12)^48     = $2540.98

tan (15°) =   [1 -cos 30 ] / sin 30  = [ 1 -sqrt (3)/2 ] / (1/2) ] =  [2 -sqrt (3)] / 1

sin (15°) =   [ 2 -sqrt 3 ] / sqrt [ 1^2  + (2 -sqrt 3)^2 ]   =  [2 -sqrt 3 ] /  sqrt [ 8 - 4sqrt 3]

cos (15°)  =  1 / sqrt [ 8 - 4sqrt 3]

tan (7.5°) =  [1 - cos 15 ] / sin 15  =  [ 1 - 1 / sqrt [8 - 4sqrt 3 ] ] / ( [ 2 -sqrt 3] / sqrt [ 8 -4sqrt 3] ) =

sqrt [ 8 -4sqrt 3 ] * [  1  - 1/ sqrt [ 8 -4sqrt 3] / [2 -sqrt 3]   =

(sqrt [ 8 - 4sqrt 3 ] [ sqrt [8 - 4sqrt 3 ] - 1 ] / sqrt [8 -4sqrt 3])  / [2 -sqrt 3]  =

(sqrt [ 8  -4sqrt 3 ] - 1)  /   [2 -sqrt 3 ]  =

[ sqrt [ 8 - 4sqrt 3] - 1 ] [ 2 + sqrt 3 ]  =    { sqrt [ 8 - 4sqrt 3 ]  = sqrt [( sqrt 6 - sqrt 2)^2] = sqrt 6 - sqrt 2 }

[(sqrt 6 - sqrt 2 )  -1 ] [ 2 + sqrt 3]  =

2sqrt 6 - 2sqrt 2 - 2 + sqrt 18 - sqrt 6 - sqrt 3   =

sqrt 6 - 2sqrt 2- 2 + 3sqrt 2 -sqrt 3 =

sqrt 6 + sqrt 2 - sqrt 3  - 2   =

sqrt (2)  + sqrt (6)  -sqrt (3)   - 2   =

sqrt (2) + sqrt (6) - sqrt (3)  - sqrt (4)

a + b =  8

c + d  =  7

C ( 19,7)  = 50388 ways

https://web2.0calc.com/questions/angle-problem_3

cos A =  24/ sqrt [ 24^2 + 7^2 ]  =  24  / 25

sin A = 7 / 25

tan (A/2)  =  [ 1 -cos A ] / sin A =  [ 1- 24/25 ] / [ 7/25 ] =  1 / 7

tan (A/2)  = CD / AC

(1/7)  = CD / 24

CD  = 24 / 7

Likewise

cos B = 7/25

sin B = 24/25

tan (B/2)  = [ 1 -cos B ] / sin B  =  [ 1 - 7/25 ] / [ 24/25]  =  18/24 = 3 / 4

tan (B/2)  = CE / BC

3/4  =  CE  / 7

CE =  21/4

Reflection over y = x just reverses the coordinates....so B = (7,5)

A baking club wants to form an executive committee. There are $20$ people in the baking club, including Mark. In how many ways can the baking club form an executive committee with $7$ people?  

This is a Combinations problem, of 20 items taken 7 at a time. 

One way this is abbreviated is             ₂₀C₇    

it's calculated as                                     20!   

                                                       –––––––––––   

                                                        (7! • (20–7)!)   

I can't write down the numbers.  

20! is too big, something times e18.  

My calculator screen won't display it all.  

The final answer works out to               77,520   

_.

\(a_n = a_{n - 1} + a_{n - 2}\)

\(a_{n-2}=a_n-a_{n-1}\)

\(a_9=4-1=3\)

\(a_8=1-3=-2\)

\(a_7=3-(-2)=5\)

\(a_6=-2-5=-7\).

7! / ( 2!)  =  2520 identifiable sequences

(x-2) (x + 6) ≤  (x - 2) ( x + 5)

x^2 + 4x - 12 ≤ x^2 + 3x -10

x ≤  2    =   ( -inf , 2 ] 

https://web2.0calc.com/questions/counting_37747

(1 + i) (2 - i) (3 + i)

( 2 + 2i - 1i - i^2) (3 + i)

(2 + i  - (-1) ) (3 + i)

( 3 + i)  (3 + i)

9 + 3i + 3i  + i^2

9 + 6i  - 1

8 + 6i

https://web2.0calc.com/questions/simplify_39168

x ( x + 6)  > 16  - x + 14 + x^2

x^2 + 6x > x^2 + 30 - x

7x > 30

x >  30 / 7    =    ( 30 / 7 , inf )

https://web2.0calc.com/questions/equilateral