Those angles have to be equal
so 3x-25=6x-80
3x-6x=-80+25
-3x=-55
x=55/3
Step-by-step explanation:
Which statements are true about the polynomial function?
f(x)=x4+5x3−x2−5x
Select each correct answer.
f(5)=0
f(x)=0
when x=−5
f(x)
divided by (x+5)
has a remainder of 0.
(x−5)
is a factor of f(x) .
When x=-5, f(x) IS 0.
f(x) divided by (x+5) has a remainder of 0. (x+5) is a factor of f(x).
But x-5 is NOT a factor of x
Answer:
Point Critical point
Q (2,0) local minimum
R (-2,1) Saddle
S (2,-1) local maximum
T ( -2,-1) Saddle
O ( -2,0) Saddle
Step-by-step explanation: INCOMPLETE ANSWER INFORMATION ABOUT THE POINTS ARE RARE
f(x,y) = x³ +y⁴ - 6x -2y² +3
df/dx = f´(x) = 3x² -6x
df/dxdx = f´´(xx) = 6x
df/dy = f´(y) = -4y
df/dydy = 4
df/dydx = df/dxdy = 0
df/dydy = f´´(yy)
D = [ df/dxdx *df/dydy] - [df/dydx]²
D = (6x)*4 - 0
D = 6*2*4 D > 0 and the second derivative on x is 6*2 = 12
so D > 0 and df/dxdx >0 there is a local minimum in P
Q(2,1)
D = (6*2)*4 D>0 and second derivative on x is 6*2
The same condition there is a minimum in Q
R ( -2,1)
D = 6*(-2)*4 = -48 D< 0 there is a saddle point in R
S (2,-1)
D = 6*2*4 = 48 D > 0 and df/dxdx = 6*-1 = -6
There is a maximum in S
T ( -2,-1)
D = 6*(-2)*(4) = -48 D<0 there is a saddle point in T
O ( -2,0)
D = 6*(-2)*4 = -48 D<0 there is a saddle point in O
I need the table and the answer choices to help you though
