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https://web2.0calc.com/questions/bisectors_25

I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row, if all three mathematics textbooks must be together?   

Temporarily consider the group of three math books as one unit. 

Then, 1 group + 2 psych + 2 chem can arrange to ₅P₅ = 120 ways. 

But wait.  You could arrange the three math books 6 different ways. 

So all the possible arrangements is 120 • 6 = 720 ways.  

_.

No prob !!!

Ax ^2 + Bx + 1

2x^2 - 5x + 6      divide this  through by 6

(1/3)x^2 - (5/6) x + 1

A + B   =  (1/3) - (5/6)  =  (2/6) - (5/6)  = -3/6  = -1/2

Thanks! The detailed explanation really helps.

4x^3 -11x^2 + 2x + 3

Test whether 3/4  is a root

3/4  [  4     -11     2      3   ]

                    3   -6      -3

   ___________________

        4       -8     -4      0

It is  a root

The remaining polynomial is

4x^2 - 8x - 4 = 0

x^2 -2x - 1  =0

x^2  -2x   =  1

x^2 -2x +1  =1 + 1

(x -1)^2  = 2       take both roots

x -1  =sqrt 2                   x - 1 = -sqrt 2

x = 1 + sqrt 2                 x = 1 - sqrt 2

These are the other two roots

Simplify as

3x^2 -kx + 33  = 0

Will have no real solutions when

k^2 - 4*3*33 < 0

k^2  - 396 < 0

k^2 < 396

k < sqrt (396)

k = 19

1  / (64a^3 + 7)  - 7 =  0

1 / (64a^3 + 7)  =  7

1 = 7 (64a^3 + 7)

1 = 448a^3 + 49

-48 = 448a^3

-48 / 448  = a^3

-3/28 = a^3

a = (-3/28)^(1/3)

480 = 2^5 * 3  * 5  = 

2 * 4 * 4 * 3 * 5 * 1 =    6! / ( 2!) =   360

6 * 4 * 4 * 5 * 1 * 1 =  6! / ( 2! * 2!)  =  180

360 + 180  =  

540 possible sequences

The probability is 3/8.

Add these

2a  =10

a = 5

5 + 5b  = 250

5b = 245

b = 49

https://web2.0calc.com/questions/geometry_38746

n! / [ (n - 2)! *2] > 400

n! / (n -2)! > 800

n * ( n -1)  > 800

n^2 - n >  800

n^2 - n - 800 = 0

n =  [1 + sqrt [1 + 3200 ] ]  / 2   ≈  28.79

n = 29

Proof 

C(28, 2) = 378

C (29, 2)  =  406

There are 4 ways.

(r -6)^2  +(r -6)^2  =r^2

2 (r -6)^2 =r^2

2[ r^2 -12r + 36 ]= r^2

r^2 - 24r + 72  = 0

r^2 - 24r = -72

r^2 - 24r + 144 = -72 + 144

(r - 12)^2  = 72             take the positive root

r - 12 = sqrt (72)

r = 12 + sqrt (72)  =  12 + 6sqrt 2  =  6 ( 2 + sqrt 2)

There is no minimum  for x > 1

The graph approaches y = 0  as x increases, but  there is no true minimum

https://web2.0calc.com/questions/geometry_89864

Let O be the cemter of the red circle and r = its radius

Let N  = the center od the right semi-circle

BN^2  + BO^2   = ON^2

1^2 + (2 -r)^2  = (1 + r)^2

1 + 4 - 4r +r^2 = r^2 + 2r + 1

5 - 4r =  2r + 1

4 = 6r

r = 4/6  = 2/3

c = 3

a(-5)^2 +b(-5)  + 3  = 0

a(4)^2  + b(4) +3  = 0

25a - 5b   =  -3   →    100a - 20b  = -12

16a + 4b =  -3   →      80a + 20b = -15

180a  = -27

a = -27/180  =  -3/20

16 (-3/20) + 4b = -3

4b  = -3 + 48/20

b = -3/4 + 48/80=   -3/20

a + b + c = 2 * ( -3/20 ) + 3  =  -6/20 + 60/20  = 54/20  =  27 / 10

                 A

         10          15

B             D               C

AD^2  = 15^2 - 3BD^2 

AD^2 = 10^2  - BD^2

     AD^2        =      AD^2      

15^2 - 3BD^2  = 10^2 - BD^2

15^2 -10^2  = 2BD^2

150 = 2BD^2

75 = BD^2

AD^2 = 10^2 - 75

AD^2 = 100 - 75

AD^2  =25

AD =  5

              A

                    8

B       2   K      3          C

AKC , AKB  are right triangles

AK = sqrt [8^2 - 3^2]  = sqrt [ 55]

AB  = sqrt [2^2 + 55] =  sqrt [59]

British Flag Theorem

AX^2 +CX^2 =BX^2 + DX^2

3^2 + 3^2  = 3^2 + DX^2

18  = 3^2  + DX^2

18 - 9  = DX^2

DX^2 =9

DX = 3

Simplify as

y = 4x^2  + 22x + 13

x coordinate of the vertex =  -22 / (2*4)  = -22/8  = -11/4

y coordinate =  4(-11/4)^2 + 22 (-11/4) + 13    =  -69/4 

Simplify as

x^2 - 5x + y^2  = 214   complete the square on x

x^2 - 5x + 25/4  + y^2  =  214 + 25/4

(x - 5/2)^2 + y^2  = 881/4

The area =   (881/4) pi ≈  691/94

ab^4 = 48

ab^2 = 4        divide the frist equation by the second

b^2 = 12

b = sqrt 12   = 2sqrt 3