Answer:
25
I hope i got it in time :)
Answer:
Step-by-step explanation:
Answer:
In parallelogram ABCD
FD is perpendicular to BC
BE is perpendicular to CD
Consider triangle BEC and triangle DFC
FC = EC (Given)
Angle BEC = Angle DFC (=90°)
Angle BCE = Angle DCF (common)
Therefore triangle BEC is congruent to triangle DFC (AAS congruency)
DF = BE (CPCT)
Since the altitudes are equal their bases will also be equal
Therefore BC = DC
Therefore BC = DC = AD = AB
Therefore ABCD is a rhombus
Hope this helps!
Mark me as brainliest
The answer is 1 over 3(3.14)(5²)(8)
The volume of the cylinder (V1) is:
V1 = π · r² · h (r - radius, h - height)
The volume of the cone (V2) is:
V2 = 1/3 π · r² · h (r - radius, h - height)
It is given:
r = 5 feet
h = 8 feet
π = 3.14
Therefore, the volume of the cone is:
V2 = 1/3 · 3.14 · 5² · 8 which is the same as 1 over 3(3.14)(5²)(8).
Answer:
C
Step-by-step explanation:
No solutions, because the graphs do not intersect.
