<span>Volume of Cone:
1/3 Bh where B is the area of the base
1/3 Bh = 5
Bh = 15
B = 15/h
The cylinder has the same base area as the cone.
Volume of Cylinder:
V = B*h
B = V/h and B = 15/h
The height of the cone and the cylinder are the same.
V/h = 15/h
Volume of Cylinder = 15.</span>
<span>3p - 6 > 21 <=> 3p > 21+6 <=> 3p > 27 <=> p > 27/3 <=> p > 9. So the answer is B. When solving inequalities the idea is to move the terms which contain the unknown quantity to one side and the rest of the terms to the other side. And then perform valid mathematical operations to find the interval value for the unknown quantity.</span>
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that 
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that 
. In this equality we can perform a right multiplication by 
and obtain 
. Then, in the obtained equality we perform a left multiplication by P and get 
. If we write 
and 
we have 
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have 
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and 
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.
<h3>Answer: 1.15</h3>
==============================================================
Work Shown:
QR = 73
QP = 55
PR = x
Pythagorean Theorem
a^2 + b^2 = c^2
(QP)^2 + (PR)^2 = (QR)^2
(55)^2 + (x)^2 = (73)^2
3025 + x^2 = 5329
x^2 = 5329-3025
x^2 = 2304
x = sqrt(2304)
x = 48
PR = 48
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tan(angle) = opposite/adjacent
tan(R) = QP/PR
tan(R) = 55/48
tan(R) = 1.1458333 approximately
tan(R) = 1.15
