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https://web2.0calc.com/questions/triangle_4602

\( a + \frac{1}{a} + a^2 + \frac{1}{a^2} = 0 \)

[a^2 + 1] / a  +  [ a^4 + 1] / a^2  =   0          multiply through by a^2

a^3 + a + a^4 + 1   =   0

a^4 + a^3  +  a + 1  = 0

a^3 ( a + 1)  + (a + 1)  =  0

(a + 1) ( a^3 + 1)  =   0

a + 1 =  0               or       a^3 + 1 =  0 

a =  -1                             a^3  =  -1

                                       a =  -1

a^5  = (-1)^5 =   -1

      A 30

12  D

      C 90                          B

Angle  B  = 60

So angle BCD =  30

Side  BC =   12/sqrt 3  =  4sqrt 3

So...in triangle  BDC,, DC  =  BC / sqrt 3 =  4sqrt 3 /sqrt 3 =  4

Area of BDC = (1/2) BC * DC =  (1/2) (4sqrt 3) (4)  =  8 sqrt 3

Area of ABC  =  (1/2) (AC)(BC)  = (1/2)(12)(4sqrt 3)  =  24sqrt 3

So....  [ ABD ]  =   [ ABC ] - [ BDC ] =  24sqrt 3 - 8sqrt 3 =  16 sqrt 3

CORRECTED 

https://web2.0calc.com/questions/triangles_35925

A

       15

          X

                 5

B                   C

ABC  ≈  BXC

ABC  ≈  AXB

So

BXC  ≈  AXB

BX / CX  =  AX / BX

BX^2  = AX * CX

BX^2 =  15 * 5

BX^2 = 75

BX =  sqrt (75)  =  5sqrt (3)

p + 2q  + 6p - 5q + 14p + 8q =  180      simplify

21p   + 5q  = 180

21p  = [ 180  - 5q ] 

p =   [ 180 - 5q ] / 21

Thae area of the triangle =  6

The slope of the line between A, B =  [2-0] / [ 3- - 5] =  2 / 8 =  1/4

The equation of this line through this segmemt  is

y = (1/4)(x + 5)   

Let the distance between  -5   and  the x we are looking for  =  x - -5 =  x + 5

And the distance between the x axis and AB  is just the eqation of our line

So....we need to solve this

(1/2) area of triangle ABC  =  (1/2)  ( x + 5) ( 1/4) (x + 5)

3 =  (1/8) ( x + 5)^2

24  = (x + 5)^2        take the positve root

sqrt (24)  = x + 5

x =  sqrt (24) - 5     

x = 2sqrt (6)  - 5    =   k 

After n steps.....he can reach ( n + 1)^2   lattice points  (points with integer coordinates)  =  (2 + 1)^2  =   9

Here's my best attempt

Let f(x)    =  y

And notice that we can write

y =  √ [ x - y ]                square both sides

y^2  =  x  - y

y^2 + y  =  x                  coplete the square on y

y^2 + y + 1/4 =   x + 1/4

(y + 1/2)^2  =  [ 4x + 1]  / 4          take the positive root

y + 1/2 =  √[4x + 1 ]  / 2

y =  ( √  [ 4x + 1 ] - 1  )   /  2

Let  x =  (n) (n + 1)

When    n           (n) (n+1)  = x              and y

= 1                            =   2                       =  1

= 2                            =   6                       =  2

= 3                           =   12                      =  3

....

= 31                        = 992                       = 31   

The largest  three-digit value for x = 992  and this produces  y = f(x)  =   31       

Let X be a point inside rectangle ABCD.  If AX = 3, BX = 3, and CX = 3, then find DX.   

                                        AX² + CX²  =  BX² + DX²         

                                         3²  +  3²  =  3²  +  DX²     

                                         DX²  =  3²  

                                          DX  =  3    

i copied the following from the internet.  

In Euclidean geometry, the British Flag Theorem says that if a point P is chosen inside a rectangle ABCD, then the sum of the squares of the Euclidean distances from P to two opposite corners of the rectangle equals the sum to the other two opposite corners. 
_.

Let "r" = the amount added to each successive term.  

a₁ + r = a₂   and   a₂ + r = a3   therefore a₃ = a₁ + r + r    

given that a₁ + a₃ = 5   therefore (a₁) + (a₁ + 2r) = 5                 (eq 1)  

a₁ + r = a₂   and   a₂ + r = a₃   and   a₃ + r = a₄   therefore a₄ = a₁ + r + r + r   

given that a₂ + a₄ = 6   therefore (a₁ + r) + (a₁ + 3r) = 6            (eq 2)   

subtract (eq 2) minus (eq 1)   

(eq 2) ...............................................  2a₁ + 4r = 6    

(eq 1) ...............................................  2a₁ + 2r = 5    

                                                                     2r = 1         

                                                                       r = 0.5      

substitute r back into (eq 2)               2a₁ + (4)(0.5) = 6   

                                                                       2a₁ = 4  

                                                                         a₁ = 2     

check answer  

                          a₁ = 2.0 , a₂ = 2.5 , a₃ = 3.0 , a₄ = 3.5    

                          a₁ +  a₃ = 5                a₂ +  a₄  = 6   

                         2.0 + 3.0 = 5               2.5 + 3.5 = 6    

If we want the system to have an infinite number of solutions, we want two congruent equations for the system. 

We have \(6a + 2b = k + 3a + mb\). We notice that if we subtract \(3a \) from both sides, the left sides of both equations match, making life much easier for us!

Subtracting \(3a \) from both sides achieves us \(3a+2b=k+mb \). In order for the equations to be equal, we want the left sides of it to match as well, meaning we can write the equation \(k+mb=2\). If m was a integer other than 0, we wouldn't have matching equations, meaning we need to have \(m = 0\). If \(m=0\), then we have \(k=2\).

Our final answer is

\(m = 0\)

\(k=2\)

Thanks! :)

Pretty easy problem in this case, so I will try my best to explain it.  

So we can choose any two binomials to multiply, and then multiply it to the third binomial. For the sake of this problem, let's first multiply \((1+i) \)with \((2-i)\). Distributing and expanding, we get that \((1+i)(2-i) = 2 - i + 2i - i^2\). Combining like terms and knowing that \(i^2=-1\), we get \(2+i+1 = 3+i\). 

Now, we multiply this by the third binomial, which coincidentally is also \(3+i\). \((3+i)(3+i) = (3+i)^2 = 8 + 6 i\)

The final answer is \(8 + 6 i\)

Thanks!

Let  
f(x) = \sqrt{x}  
Find the largest three-digit value of x such that f(x) is an integer.  

31²  is    961   

32²  is  1024   

Therefore, x = 961   

_.

Evaluate the dicrimimant of each......the one that =  0  will be the one with the repeated root

Let the continued fraction   =  x

So we have

x  =        1 

        ______

        1 +  1 

             ____                      simplify

             4 + x

x =     1

      ________

      [4 + x + 1]

    __________

          [ 4 + x  ]

x =  [ x + 4 ]  /  [ x + 5]

x^2 + 5x  = x + 4

x^2   + 4x  =   4

x^2 + 4x + 4  =  4 + 4

(x +2)^2  =   8        take the positive root

x + 2 = 2sqrt (2)

x = 2sqrt 2 - 2  =  2 ( sqrt 2  - 1)

Multiplying each fraction  by its conjugate in the numerator and  denominator we get

[ sqrt 36 - sqrt 39 + sqrt 39 - sqrt 42 + sqrt42 -sqrt 45 +   .....+ sqrt 93 -  sqrt 96 + sqrt 96 - sqrt 99 ]  /  -3    =

{ All terms  "cancel" except  those in  red }  =  

[ sqrt 36 - sqrt 99 ] / -3   =

[sqrt 99 - sqrt 36 ] /  3   =

[ 3sqrt 11 - 6 ] / 3  = 

sqrt 11 -  2

a11 - a1  =   4

a10 + a9  = 4 + a1

(a8 + a9) + a9  = 4 + (a3 - a2)

2a9 + a8  = 4 + a3 - a2

2(a8 + a7) + a8  = 4 + ( a5 -a4) - ( a4 - a3)

3a8 + 2a7  =   4 + a5 - 2a4 + a3

3 ( a7 + a6) + 2a7  = 4 + (a7 -a6) - 2(a6 -a5) + (a5 - a4)

5a7 + 3a6 =  4 + a7 - 3a6 + 3a5 - a4

4a7 + 6a6 = 4 + 3a5 - a4

4a7 + 6a6  = 4 + 3(a7 -a6) - (a6 -a5)

4a7 + 6a6  = 4 + 3a7 - 4a6 + a5

a7 +10a6 = 4 + a5

(a6 + a5) + 10a6 = 4 + a5

a6 + a5  + 10a6 = 4 + a5

11a6  = 4

a6 =  4 /  11

a - 1/a  = 4        square both sides

a^2 - 2 a(1/a) + 1/a^2  = 16

a^2 - 2 + 1/a^2  =  16

a^2 + 1/a^2  =  18

a^3 - 1/a^3 =   

( a  - 1/a)  ( a^2 + a(1/a) + 1/a^2)   =

(4) ( a^2 + 1/a^2 + 1)  =

(4) ( 18 + 1)   =

76

We can solve this problem by considering the different cases for how many candies the twins can get:

Case 1: Twins get 0 candies each

In this case, we essentially have 3 children (excluding the twins) who can receive 3 identical candies.

Distributing 3 candies to 3 children can be done in (3 + 3 - 1)! / (3 - 1)! = 6 ways using the stars and bars method (where 3 stars represent candies and 2 bars represent dividers separating the children).

Case 2: Twins get 1 candy each

Now, we have to distribute 1 candy to each twin (which is fixed) and 1 remaining candy to the 3 remaining children.

Distributing 1 candy to 3 children can be done in 3 ways.

Case 3: Twins get 2 candies each

Here, the twins get a total of 4 candies (2 each), leaving 1 candy for the remaining child.

There's only 1 way to distribute this remaining candy.

Total Ways

To find the total number of ways to distribute the candies, we add the number of ways for each case:

Total Ways = Cases (Twins get 0) + Cases (Twins get 1) + Cases (Twins get 2)

Total Ways = 6 + 3 + 1 = 10

Therefore, there are 10 ways to distribute the 3 candies such that the twins get an equal amount.

Here is the answer:

The infinite geometric series with first term  3  and common ratio  4   has a sum of    15.