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Let's  find the point (P) of attachment....we have that

sin (36°)    =   P  / 60    multiply both sides  by  60

60sin (36°)  = P

So....in the second situation we have  that

sin (new angle)    = 60sin(36°) /  110

sin (new angle)  =  6sin (sin 36°)  /11

Take the arcsin

arsin  [ 6 sin (36°) / 11]   =  new angle ≈  18.7°

LOOKS GOOD, JENNY     !!!!!

Thanx for the correction , Geno!

See my corrected answer  under your original post, qwertyzz

sqrt 40 because the difference between 5 and 8 is 3. Using Pythagorean Theorem we get 7^2=3^2+x^2. assuming x is the missing length, we get 40=x^2. Therefore it is sqrt 40.

I asked the Question right after this. Can someone please answer it!

I do not know why there is a limit.....  there is not one for registered users.   Just open an account.

I can't read that.  Maybe do a screenshot instead?

                 C

       14

D              9                      E   =  71°

We  can apply  the Law  of Sines  to  find the measure  of  angle  DCE....we  have  that

sin DCE / 9  = sin CED  /  14

sin DCE   / 9  =   sin (71)  / 14

sin DCE  =  9 sin (71)  /14

sin DCE  = .6078

Taking the arcsin  we have that

arcsin (.6078)  =  DCE  ≈ 37.43°

So   angle CDE    = 180  - 71  - 37.43   =  71.57° = 71.6° 

No answerer is paid on here, they are just kind people willing to help out others. 
 

 A circle is centered at (5,15) and has a radius of √130 units. Point Q = (x,y) is on the circle, has integer coordinates, and the value of the x-coordinate is twice the value of the y-coordinate. What is the maximum possible value for x? 

We  have  this equation

(x - 5)^2 + (y - 15)^2  =  130

Let the point we seek  be  ( 2y, y)     

So we have

(2y -5)^2   + ( y - 15)^2  =  130        simplify

4y^2  - 20y + 25  + y^2  - 30y + 225   =  130

5y^2   - 50y + 250  = 130

5y^2  - 50y +  120  =  0        divide through  by  5

y^2  - 10y  + 24  = 0            factor

(y - 6) ( y - 4)  =  0

Setting each factor to 0   and solving for   we  get that

y  = 6                  or              y = 4

So...it's obvious that x is maximized when  y =  6....and x  = 12

See the graph here : https://www.desmos.com/calculator/bkoknegsvd

But like why? Do they get paid or something?  I was in the middle of helping some people! It literally says no registration required?!?!?!?!?!?

Because they don't want a whole bunch of guests they want members.  The guest option is so you can try out the website and see if you want to become a member.

2. The graphs of y=|x| and y=-x^2-3x-2 are drawn. For every x, a vertical segment connecting these two graphs can be drawn as well. Find the smallest possible length of one of these vertical segments.

Look at  the graph here : https://www.desmos.com/calculator/p7iq0yf8z5

Note  that  the  vertex  of  the parabola  occurs at  (-1.5, .25)

When a  vetical ine is drawn at x  =  -1.5, the line will intersect the absolute value graph at (-1.5, 1.5)

So......the distance between  these two points is

1.5 -  .25   =   

1.25  units

Similiar Problem:

Figure out what you have and what you need, then set up the equation.

You have 870 dollars total for all seats sold:

You have main floor seats sold at 11 dollars a piece.

You have balcony seats sold at 7 dollars a piece.

Now you need to know how many of each were sold.

You have 2 times as many floor seats sold than balcony seats, so we say:

Let x = balcony seats sold

Let 2x = floor seats sold

Set up the equation:

x(7) + 2x(11) = 870

7x + 22x = 870

29x = 870

x = 30 ; This is the amount of balcony seats:

2x = 60 ; This is the amount of floor seats:

sin²(x) + sin(x) - 2  =  [ sin(x) + 2 ][ sin(x) - 1 ]

Assuming that ABCD is a trapezoid.

Drop a perpendicular from A to BC; call this point X.

Triangle(BAX) is a 30^o - 60^o - 90^o right triangle with AB the hypotenuse.

To find AX, divide AB by the sqrt(3)   --->   AX  = 6·sqrt(3)/sqrt(3)  =  6  [this is the height of the trapezoid]

BX  =  6·sqrt(3)/2  =  3·sqrt(3).

Drop a perpendicular from D to BC; call this point Y.

XY  =  AD  =  16.

Triangle(DCY) is a 30^o - 60^o - 90^o right triangle with CD the hypotenuse.

DY  =  AX  = 6.

YC  =  DY/sqrt(3)   --->   YC  =  6/sqrt(3)  =  2·sqrt(3).

BC  =  BX + XY + YC.

do you have any other info that might help us solve the problem?

7-2=5