CPhill

a3 = 3a1 + 2d

a4 = 4a1 + 3d

a6 = 6a1 + 5d

a1 + a3  =  -2

a4 + a6 = 1

So

a1 + 3a1 + 2d =  -2

4a1 + 3d + 6a1 + 5d = 1

So

4a1 + 2d  = -2  →    -16a1 - 8d = -8   (1)

10a1 +8d = 1     (2)

Add (1) , (2)

-6a1 =  -7

a1  = 7/6

Let a1 be the first term and d the common difference between rwems

5a1 + 4d =  1/5   (mult through by -2)   →   -10a1 -8d = -2/5   (1)

10a1 + 9d  = 1/10     (2)

Add (1) , (2)

d = -3/10

5a1 - 4(3/10)  = 1/5

5a1 -12/10 = 1/5

5a1 - 6/5 = 1/5

5a1 = 7/5

a1 = 7/35

S15  =   15a + 14d  =  15(7/35) + 14 (-3/10)  =  -6/5

7! / 2!  =   2520 different sequences

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

Law of Cosines

BC^2  = AC^2 + AB^2  - 2 (AC * AB) cos (BAC)

7^2  = 5^2 + 3^2 - 2 ( 5 * 3) cos (BAC)

[ 7^2 -5^2 -3^2] / [ -2 * 5 * 3] =  cos (BAC)   =  -1/2

arccos (-1/2)  =  BAC =   120°

Since AD is a bisector ofthis angle, then BAD  = 60°

cos 60°  = cos (BAD)  =  1/2

side AB  is contained on the line

y = (3/5)x

side AC is  contained on the line

y = -(1/4)(x -32)  = (-1/4)x + 8

The x intersection of these lines is

(3/5)x  = -(1/4)x + 8

[ 3/5 + 1/4] x  = 8

[17/20]x =8

x = 8 *20 / 17  = 160 /17

And the y value at this  point =  (3/5)(160 /17)  = 96/17  = the altitude...call this AD

[ABC]  =  (1/2) BC * AD   =  (1/2)  32 * 96/17 =  1536  /  17  

Complete the square on x, y

x^2 -10x + 25 + y^2 + 10y + 25  =  25 + 25 - c

(x -5)^2  + ( y+ 5)^2  =  50   - c

We want 50 - c  =1

c =  49

We have 6 equilatera ltriangles with sides  =3

The area =    6 (1/2) *3^2 * sqrt 3  /2 =  (27/2)sqrt 3

https://web2.0calc.com/questions/geometry_74033

When Dirk finishes, Edith has run 14 km and Foley has run 9 km

So ....  his rate to  her rate =  9/14

When  she finishes, Foley has run   (9/14 )* 24 km

So he is   24 -  (9/14) * 24  =  24 ( 1 - 9/14)  =  24 * ( 5/14)  ≈  8.57 km behind