Angle P is greater than angles Q and R.....therefore.....
The exterior angle to P will have the smallest degree measure.
No prob !!!
{Well, I guess there were a couple of them....LOL!!! }
Well
2*8 = B = 16
And
y =Ax^2 + B .....so....
8 = A(2)^2 + 16
-8 = 4A
A = -8/4 = -2
(A,B) = (-2, 16)
Oops!!!...I was in a hurry !!
Corrected !!!
This changes the answer to the other problem, too....will correct !!!
Believe it, or not....we have a lower power polynomial over a higher power polynomial
When this happens, we have a horizontal asymptote at y = 0
So
The range is (-inf ,0) U ( 0 , inf)
See asinus's answer here :
https://web2.0calc.com/questions/geometry_94201
Mmmm....I don't know !!!
Can you try making a new account ???
I don't have any control over the accounts
The denominator simplifies to
2y -10
Since this cannot = 0 then y = 5 is NOT in the domain
domain = (-inf , 5) U ( 5 ,inf )
X
28
Z Y
cos Y = ZY / XY
7/29 = ZY / 28
(28 * 7) / 29 = ZY = 196/ 29
XZ = sqrt (28^2 - (196/29)^2) = (168)sqrt (22) / 29
Tan X = ZY / XZ = (196/29) / [ 168 sqrt (22) / 29] = 196 / [168 sqrt (22)] =
7/ [6sqrt (22) ] =
7 sqrt 22 / ( 6 * 22) =
7sqrt (22) / 138
No denominator can = 0 .... so.....
x - 64 = 0
x =64 is not in the domain
x^2 - 64 =0
x^2 =64
x = -8 or x =8 are not in the domain
x^3 - 64 = 0
x^3 =64
x =4 is not in the domain
