To solve the question we shall use the formula for the range given by:
Horizontal range, R=[v²sin 2θ]/g
plugging in our values we get:
500=[160²×sin 2θ]/10
5000=160²×sin 2θ
0.1953=sin 2θ
thus:
arcsin 0.1953=2θ
11.263=2θ
hence:
θ=5.6315°~5.63
F(x)=3x²-6x+13
a=3, b=-6, c=13
the x coordinate of the vertex is x=-b/(2a), so x=-(-6)/(2*3)=1
when x=1, y=3(1)²-6(1)+13=10
the vertex is at (1,10)
answers are in bold.
Factor the numerator to get:
(-3z(z+5))/(z+5) now you notice that (x+5) cancel out leaving:
-3z
The law of cosines is
c= square root of a^2 + b^2 - 2ab cos c
the law of sines is
a = b(sin a /sin b)
Answer:
x<-1
Step-by-step explanation:
-4x+8>12
-4x>12-8
-4x>4
x>4/-4
x>-1
x<-1
