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makvit [3.9K]
3 years ago
10

Can you please help me with this, I have no idea

Mathematics
1 answer:
sergejj [24]3 years ago
6 0
I have no clue sorry
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Solve and check:  3 x2 5x 6 x − 1 x 2 = 7 x 3
oee [108]

Answer:

x =  7 is a solution.

x = -2 is an extraneous solution .

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3 years ago
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Crystal earns $5.50 per hours mowing lawns. Write a rule to describe how the amout of money m earned is a function of the number
ra1l [238]
She earns $5.50 for every hour so:
m = 5.5h
3 hours and 45 minutes is 3.75 hours.
m = 5.5(3.75)
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After rounding, she earns $20.63.
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There are 45 new houses being built in a neighborhood. Last month 1/3 of them were sold. This month 2/5 of the remaining houses
polet [3.4K]

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total=45

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=4/15

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12.09 rounded to the first decimal place
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Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

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.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
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