3SQRT(3). It is equal to the short side time Square Root of 3.
The problem is asking for the height of the fluid using the following information:
Volume of fluid
=
V
=
36
π
cm
3
Radius of new cylinder
=
r
=
3
cm
To start the solution, solve for the height
h
in the formula for the volume of a cylinder:
V
=
π
r
2
h
π
r
2
h
π
r
2
=
V
π
r
2
h
=
V
π
r
2
V =π
r
2
h
π
r
2
h
π
r
2
=
V
π
r
2
h =
V
π
r
2
Substituting the values of the volume and the radius, the height of the fluid is:
h
=
V
π
r
2
=
36
π
cm
3
(
3
cm
)
2
=
(
4
)
(
9
)
π
cm
3
(
9
)
cm
2
=
4
π
cm
=
12.5663706144
≈
12.6
cm
h =
V
π
r
2
=
36
π
cm
3
(
3
cm
)
2
=
(
4
)
(
9
)
π
cm
3
(
9
)
cm
2
=4π cm =12.5663706144≈12.6 cm
Thus, the fluid reaches up to
12.6
cm
12.6
cm
in the new cylinder.
Answer:
2x + 3y = -6
-4x + 3y = 12
5x + 3y = }
x - y = 5
Step-by-step explanation:
Standard Form is 
.
2x + 3y = -6, -4x +3y = 12, 5x + 3y = '?', and x - y = 5 are the equations that are in standard form.
'2x+3 = 6' does not have the 'y' to make it standard form.
'y = 2x + 5' is slope-intercept form.
Answer:
angles KKL & FGE
Step-by-step explanation:
Alternate Exterior Angles (AEA) are a pair of angleson the outer side of each of those two lines but on opposite sides of the transversal.
Thus for this question the AEA are:
angles KJL & FGE
