See my answer, Jenny, under your original post....not too sure about it, though !!!
Here's my best attempt
Mathematical expectation =
Winnings * probability of winning - cost to play
5(.08) + 6(.26) + 7(.34) + 9(.20) + 12(.12) - 8 = -21/30 = -0.42 teeth
Oops small mistake...corrected !!!
No prob .....
The vertex is at (-1, -2)
We have the form
y = a ( x - - 1)^2 - 2
y = a ( x + 1)^2 - 2
And the point 0,-1) is on the graph...so we can find a
-1 = a ( 0 + 1)^2 - 2
-1 + 2 = a
1 = a
So we havr
y = 1 ( x + 1)^2 - 2
y = x^2 + 2x + 1 - 2
y = x^2 + 2x - 1
a = 1 b = 2 c = -1
a + b + c = 1 + 2 - 1 = 2
The vertex is (2,1)
And we have the form
y = a(x - h)^2 + k where ( h, k) is the vertex
And the point (0, 5) is on the graph....so...
5 = a ( 0 - 2)^2 + 1
5 - 1 = 4a
4 = 4a
a = 1
So we have
y= 1( x - 2)^2 + 1
y = x^2 - 4x + 4 + 1
y = x^2 - 4x + 5
a = 1 b = -4 c = 5
a + b + c = 1 - 4 + 5 = 2
Good job, Jenny !!!
SR = ST
So angles SRT and STR are equal
So angle STR + angle UTS = 180
So
2x - 60 + 8x = 180
10x - 60 = 180
10x = 180 + 60
10x = 240
x = 240 / 10 = 24
distance between ( -2 , -2sqrt (2) ) and ( 1 , sqrt (2) ) =
sqrt [ ( 1 - -2)^2 + ( sqrt 2 - - 2 sqrt 2)^2 ] =
sqrt [ 3^2 + ( 3sqrt (2)) ^2 ] =
sqrt [ 9 + 18 ] = sqrt (27)
Area = pi (sqrt (27) )^2 = 27 pi
Triangles QTS and QRS are congruent
So angles QST abd QSR are equal....so....
4x + 5 = 6x -13
13 + 5 = 6x - 4x
18 = 2x
x = 18/2 = 9
Angle RSQ = 6(9) - 13 = 54 -13 = 41°
