Answer:
Step-by-step explanation:
-p(d+z)=-2z+59
-pd-pz=-2z+59
lets bring all z integers on one side
-2z+pz = -pd-59
z(-2+p)=-pd-59
z = (pd-59) / (-2+p)
A right cylinder has a radius of 3 and a height of 12. What is its surface area?
1 answer:
Answer:
90π units²
Step-by-step explanation:
(refer to attached)
Total Surface are off a right cylinder = area of its ends + area of curved surface
Given radius, r = 3 units and height, h = 12 units
Area of 2 ends,
= Area of 2 circles
= 2 x πr²
= 2π (3²)
= 2π (9)
=18π units²
Area of curved surface,
= 2πrh
= 2π(3)(12)
= 72π units²
Hence,
Surface area = 18π + 72π = 90π units²
You might be interested in
We have been given a function . We are asked to find the formula which gives the x-coordinates of the maximum values for given function.
We know that cosine function oscillates between and 1 that is minimum value of cosine is
and maximum value is
.
We also know that and after period of cosine is
. This means after every
, we will get to
.
So maximum of cosine is , where n is an integer.
Therefore, our required formula is , where,