- Diameter of cylinder is <u>1</u><u>4</u><u> </u><u>units.</u>
<h3><u>Explamation </u><u>:</u></h3>
<em><u>Given </u></em><em><u>:</u></em><em><u>-</u></em>
- Volume of cylinder = 245π cubic units
- Height of cylinder = 5 units
<em><u>To </u></em><em><u>Find </u></em><em><u>:</u></em><em><u>-</u></em>
<em><u>Solution </u></em><em><u>:</u></em><em><u>-</u></em>
<em>Firstly </em><em>lets </em><em>calculate </em><em>radius </em><em>of </em><em>cylinder </em><em>by </em><em>using </em><em>formula </em><em>of </em><em>volume </em><em>of </em><em>cylinder,</em><em> </em><em>as </em><em>we </em><em>know </em><em>that;</em>
- Volume of cylinder = πr²h
<em>Putting </em><em>all </em><em>values </em><em>we </em><em>get;</em>
➸ 245π = π × r² × 5
<em>By </em><em>cutting </em><em>'π' </em><em>with </em><em>'π' </em><em>we </em><em>get;</em>
➸ 245 = r² × 5
➸ 245/5 = r²
➸ 49 = r²
➸ √(49) = r²
➸ √(<u>7</u><u> </u><u>×</u><u> </u><u>7</u><u>)</u> = r²
➸ 7 = r
➸ r = 7 units
- <u>Hence,</u><u> </u><u>radius </u><u>of </u><u>cylinder </u><u>is </u><u>7</u><u> </u><u>units.</u>
<em>Now </em><em>lets </em><em>calculate </em><em>its </em><em>diameter,</em><em> </em><em>as </em><em>we </em><em>know </em><em>that;</em>
<em>Putting </em><em>all </em><em>values </em><em>we </em><em>get;</em>
➸ Diameter = 7 × 2
➸ Diameter = 14 units
- <u>Hence,</u><u> </u><u>diameter </u><u>of </u><u>cylinder </u><u>is </u><u>1</u><u>4</u><u> </u><u>units.</u>
The composed fraction
means that the input of
is the ouput of
.
So, the chain is

The domain of
is composed by all inputs which are not zero, otherwise we would have a zero denominator.
So, since
doesn't accept 0 as an input, and we want to feed it with
, we conclude that
can't be zero.
So, we have

Since

we have

Since these two points cases
to vanish, they can't be accepted as inputs by
.
total, t = 15/16
red beans, r = 3/16
pinto beans, p = 5/16
black beans, b = t -(p+r) = 7/16
The graph is attached.
Answer:
(-2.2, 4) and (8.2, 4)
Step-by-step explanation:
In an ellipse, there is a minor radius and a major radius.
Let major radius be = a
Let minor radius be= b
From the graph, we are given:
Major radius, a = 6
Minor radius, b = 3
Now, let's find the distance from the center to the focus using the formula:

Substituting values, we have:



≈ 5.2
We can see from the graph that center coordinate is (3, 4). Therefore, the approximate locations of the foci of the ellipse would be:
(3-5.2, 4) and (3+5.2, 4)
= (-2.2, 4) and (8.2, 4)