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No it means that they don't know what to say.

Angle DEF = 48 degrees.

hello

a + b + c = 1 + 6 + 2 = 9.

dy/dx = -2ysin(x/y) + 1

Let  the  number  that  Jack  gave  away  = G

Then Luis  gave away  .25G

Number  that  Jack  had  remaining =    G +  40

So Jack  must  have  started  with  G + G + 40   =    2G  +  40

And Luis   must  have started  with  (.25)G  + .25(G + 40)   =  .5G  + 10

So

(2G + 40)  +  (.5G  + 10) =  350

2.5G  +  50  =   350

2.5G  = 350  -  50

2.5G  =  300

G = 300  / 2.5 

G = 120

Jack  had   120  +  40  =  160  at  the  end

Proof

Jack   started  with    2(120) + 40  =  280 

Luis  started  with .5(120)  +  10   =   70

Jack  gave  away  120   and ended  up with  160  = 40  more  than he gave  away

Luis gave  away  (.25)(120) =  30

And  he ended up with  40   =  .25 of (160)

                                       707,106,781

cos∠ABM = cos∠(45º) = --------------------   

                                     1,000,000,000

Find   the  area

S = the  semi-perimeter =  [ 17 + 17 + 18 ]  / 2  =  26

Area =   sqrt [ 26  (26 - 17)^2 (26 - 18)  ]  =   36sqrt (13)

Area  =  (1/2)  ( perimeter)  *  inradius

36sqr (13)  = (1/2) (52) * inradius

36 sqrt (13)  = 26  *  inradius

(36/26) sqrt (13)  = inradius  =   (18 sqrt (13)  )  /  13  ≈ 4.99

Evaluate the expression.
(–16– (–10)÷(–15+12)×13   

                                – 6                    2

                             ————    =      ——  

                             (–3)(13)              13    

_.

Oh okay so I got \(\sqrt{4}\) which is also 2

The point (5,8) must be on the graph.

Lol okay.

THX....I usually  see it  better if I can assign coordinates  in  a problem .....

OK...we'll  keep our fingers  crossed....LOL!!!

I like how you assigned coordinates to each of the vertices! This is very useful for coordinate proofs! I should use that in the future

Let  the point  at  the  top of  the left pole =  (0,15)

Let  the point  where  the  tightrope  walker is  =   ( 4, 3)   

Let  the point on the  top right pole = (14,15)

Using  the distance  formula.....the  rope  length is

sqrt  [ ( 4^2  +(15 - 3)^2 ]  +  sqrt  [ ( 14 -4 )^2  + (15 - 3)^2 ]  = 

sqrt [  16  + 12^2]  +  sqrt [ 10^2  + 12^2]  = 

sqrt [ 160  ] +  sqrt [ 244  ]   ≈  28.27  m

The triangle with the side length 4 also has a height of 15-3, which is 12. Using the Pythagorean theorem, the hypotenuse is 4sqrt10. 

On the other hand, we can split the quadrilateral on the left into a rectangle and a triangle, and that triangle has a base of 14-4=10 and a height of 15-3=12. Using the Pythagorean theorem, we get that the hypotenuse of that triangle is 2sqrt61.

Therefore, the length of the rope is $2\sqrt{61}+4\sqrt{10}$

See diagram below:

  Use pythagorean theorem to calculate ?    and ??  below

   sqrt ( 12^2 + 4^2)    +  sqrt ( 12^2 + 10^2) = rope length