How to find the range of f(t)=t^2-6t

The range is the highest y value to the lowest y value

ok

find the vertex because it is the highest or lowest point in the equation
make into f(t)=a(x-h)²+k form
(h,k) is the vertex
k is the y value of the vertex
if a>0, then the vertex is the minimum
if a<0, then the vertex is the maximum

so
f(t)=t²-6t
f(t)=t²-6t+9-9
f(t)=(t-3)²-9
f(t)=1(t-3)-9

k=-9
a=1
1>0
so y=-9 is the minimum value
the maximum value will be infinity

so the range is from y=-9 to infintiy
so R=[9,∞)

101.08 degrees to nearest hundredth.

Step-by-step explanation:

Use the Cosine Rule:-

25.7^2 = 10.7^2 + 21.4^2 - 2*10.7*21.4 * cos x   where x is the required angle.

cos x =     (25.7^2 - 10.7^2 - 21.4^2) / (-2*10.7*21,4)

=  88.04 / - 457.96

= -0.192244

x = 101.08 degrees