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Hunter-Best [27]
3 years ago
13

A number is six greater than 12 times another number the sum of the two numbers is 214 find both numbers

Mathematics
1 answer:
stich3 [128]3 years ago
6 0
Let x and y represent the two numbers.
.. x = 6 +12y . . . . . one number is 6 greater than 12 times another
.. x + y = 214 . . . . their total is 214

Substitute for x in the second equation using the first equation.
.. (6 +12y) +y = 214
.. 13y = 208 . . . . . . . . simplify and subtract 6
.. y = 16 . . . . . . . . . . . divide by 13
.. x = 214 -y = 198

The numbers are 198 and 16.

_____
These are solved easily any of several ways on a graphing calculator. You should learn to use yours for that purpose.
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Step-by-step explanation:

<em>the black lines would mean where the fractions and decimals are</em>

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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 A=J^{-1}AJ. 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 A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. 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 A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

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A. $20,950.75<br><br> B.$28,458.50<br><br> C.$31,045.50<br><br> D.$34,650.75
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Answer:

C) 31,045.50

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Therefor, the CEO made $31,045.50 that month.

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