CPhill

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                             A

                                              12

        12   Z                        Y

                                                    6

B                                                    C

Since BY is a bisector

AB / AY  = BC / CY

12 / 12  = BC / 6

BC = 6

And since CZ is a bisector  let BZ = x   and AZ   = 12  - x

BZ / BC = AZ / AY

x / 6  = (12 - x) / 12

12 x  = 6 (12 - x)

12x = 72  - 6x

18x  = 72

x = 72/ 18  = 4 =  AZ

x = AC  / sqrt 2  =  2 / sqrt 2  =  sqrt 2

https://web2.0calc.com/questions/sequence_48052

4 h 36 min =   4 + 36/60  =  4.6 hrs

Let the speed of the plane in still air = S

Let the speed of the plane against the wind = S  - 40

Let the speed of the plane with the wind = S + 40

Let the distance one way = D.... so round trip =  = 2D

Distance / Rate  =  Time

We have (with no wind)

2D / S  = 4.6

2D = 4.6S

D = 2.3S

With wind

D / (S + 40)  + D / (S -40)  = 5

D (S-40) + D (S + 40) = 5 (S + 40) (S - 40)

2DS  =  5 ( S^2  - 1600)

2 (2.3)S * S  =  5S^2  -  8000

4.6S^2  = 5S^2  - 8000

.4S^2  = 8000

S^2  =  (8000) / .4 =   80000/4  =  20000

S  =  sqrt (20000) =  141.42

D =  2.3 (141.42) ≈  325.3 miles 

If three of the history books are together

They can occupy  positions 123 and  be arranged in 3! ways.....the other history book  can occupy positions 5, 6 or7

And the other three math books can  be arranged in 3! ways

So   3! * 3 * 3!   =  108 ways

They can occupy  positions  234 and  be  arranged in 3! ways.....the oher history book can occupy positions 6 or 7

And the three math books can  be arranged in 3! ways

So   3! * 2 * 3!  =  72 ways

And occupying positions  345 or 456   also  produce 2 ( 72)  arrangements   =144

And occupying  567   also produces another 108  arrangements

So  if three of the history books are together we have  2(108) +3(72)  = 432 arrangements

If all 4 of the history books  are together they can occupy positions 

1 - 4

2 - 5

3 - 6

4 - 7

For each of these, they can  be arranged in 4! ways  and the other 3 math books can be arranged in 3! ways

So....if all are together, they can  be arranged in  4 * 4! * 3!   =  3456 ways

So....the total number of arrangements =  432 + 3456  =  3888 total arrangements

7!   =  5040  ways

  30    A

                     X

90     B       12             C

Angle B = 60

AB =  12sqrt 3

AC  = 24

Since BX is a bisector...let  XC  = 24-m   and AX = m

And we have that

m / AB  = (24-m) / BC

m / (sqrt 3)  = (24 -m)

m = (24 - m) sqrt 3

m = 24sqrt 3 - msqrt 3

m+ msqrt 3 = 24sqrt 3

m ( 1 + sqrt 3)  =24sqrt 3

m = 24sqrt 3 / ( 1 + sqrt 3)  =  12sqrt3 ( sqrt 3 - 1)  = 12 [ 3 - sqrt 3 ]

[ BXA ]  = (1/2) AX * AB * sin 30°  = (1/4) AX * AB  =  12 [3 -sqrt 3 ] * [12 sqrt 3] / 4 =

3 [ 3 -sqrt 3 ] [ 12sqrt 3 ]  =  108 (sqrt 3 - 1)

                                     Q

                        M        

                                    O  

                                    1

           P                       N                        R

                                     1

If QN perp to PR  then QNR forms a right triangle  such that

ON^2 + (PR/2)^2  =  OR^2

1^2 +  (1/2)^2  = OR^2

1 + 1/4   =  OR^2

5/4  = OR^2

QR =  sqrt (5)  /  2

https://web2.0calc.com/questions/triangle_41241