Ok so
what we do is
how much greater=wasted space=cylindervolume-conevolume
volume of a cinlinder=hpir^2
volume of a cone=(1/3)hpir^2
we can alread sybsitute something
(3/3)hpir^2-(1/3)hpir^2=(2/3)hpir^2
so wasted space=(2/3)hpir^2
diameter=4
d=2r
d/2=r
4/2=2=r
and 6=height
so
2/3(6)(3.14)(2^2)
4(3.14)(4)
16(3.14)
50.24 in^3 wasted
Suppose we have a generic equation of the form:
1) The first thing you must do to complete squares is to group similar terms of the form:
2) The second step is to complete the squares.
3) The third step is factor the quadratic equation
4) The fourth step is to identify the center and the radius
Answer:
the first step in finding the center of the circle by completing the square is:
C group like terms
Answer:
w = 
Step-by-step explanation:
Sin(60) =
/ w
Sin(60) * w = 
w =
/ Sin(60)
w = 
Hope this helps!
Answer:
Line JK passes through points J(-4,-5) and K(-6, 3).
Line JK has form of y = mx + b
=> -5 = -4m + b
=> 3 = -6m + b
Subtract 2nd equation from 1st equation:
=> -8 = 2m
=> m = -4
Substitute back m = -4 into 1st equation:
=> -5 = -4(-4) + b
=> b = -21
Hope this helps!
:)