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Artist 52 [7]
3 years ago
13

Anyone good at math??

Mathematics
1 answer:
aalyn [17]3 years ago
3 0
I am what can i help you with?

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Its complete solution <br>and answer is 45​
lora16 [44]

Answer:

8

Step-by-step explanation:

6 0
3 years ago
If you could help I would really appreciate it, but if not that’s fine. Thank you.
Kobotan [32]
Since 7 is greater than 5, you input it into the last equation.

2•(7) -7
14-7
7

The answer is 7!
3 0
2 years ago
Find the additive inverse of 7/10
11111nata11111 [884]

\huge\fbox{Hi~there!}

Remember, a number's additive inverse is simply its opposite.

Let's say we have a number a.

The opposite of a is -a, and the opposite of -a is a.

Thus, the additive inverse of

\displaystyle\frac{7}{10} is

\displaystyle-\frac{7}{10}

Hope it helps.

~Just a felicitous girl

#HaveAnAmazingDay

Feel free to ask if you have any doubts.

\bf{-MistySparkles^**^*

5 0
2 years ago
Sydney purchases a new car for $27,500. The value of the car decreases at rate of 15% each year. In approximately will the $1,25
Pani-rosa [81]

Answer:

Time taken n = 19 years (Approx)

Step-by-step explanation:

Given:

Amount of car P = $27,500

Decrease rate r = 15% = 0.15

Final amount A = $1,254

Find:

Time taken n

Computation:

A = P[1-r]ⁿ

1,254 = 27,500[1-0.15]ⁿ

0.0456 = [0.85]ⁿ

Time taken n = 19 years (Approx)

8 0
2 years ago
Read 2 more answers
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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