We are given the points
<span>(0,-4),(1,0),(2,2)
The standard form a quadratic function (in terms of x) is
y = Ax2 + Bx + C
Subsitute the points
-4 = 0 + 0 + C
C = -4
0 = A + B - 4
2 = 4A + 2B - 4
Solve for A and B
A = -1
B = 5
The function is
y = -x2 + 5x - 4</span>
This is the concept of algebra, to solve the expression we proceed as follows;
cos 2x-cosx=0
cos 2x=cosx
but:
cos 2x+1=2(cos^2x)
thereore;
from:
cos 2x=cos x
adding 1 on both sides we get:
cos 2x+1=cos x+1
2(cos^2x)=cosx+1
suppose;
cos x=a
thus;
2a^2=a+1
a^2-1/2a-1/2=0
solving the above quadratic we get:
a=-0.5 and a=1
when a=-0.5
cosx=-0.5
x=120=2/3π
when x=1
cos x=1
x=0
the answer is:
x=0 or x=2/3π
Ohyeah free points bruhoo
〽Hola User_______________
⭐Here is Your Answer..!!!
______________________
↪Cartesian Coordinates of the system....!!
↪X = rcos⊙ and Y = rsin⊙
↪since here Radius (r) = 10 units and Theta (⊙) = 225°
↪there fore substituting we get as ..
↪X = 10 ( cos 225° ) = 10 * -1/ root 2 = -7.071
↪Y = 10 ( sin 225° ) = 10 * -1/root2 = = -7.071
↪the cordinates are as follows
↪(x,y) = (-7.071, -7.071 )
Starting from what I know (the difference between ribbons), I decided to go from the bottom and work my way up. I then noticed a pattern (each sum was three more of the previous one), and decided to keep my pattern of the three numbers but not have to do any major mental work and instead add three to the previous sum until I got to 38.
