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sergij07 [2.7K]
3 years ago
7

Sebastián está haciendo un curso de patinaje. El primer día recorrió una distancia de

Mathematics
1 answer:
jeyben [28]3 years ago
8 0
Sorry but I don’t understand your language.
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If two matrices, A and B, are equal, which of the
alexandr402 [8]

If A and B are equal:

Matrix A must be a diagonal matrix: FALSE.

We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:

A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]

Both matrices must be square: FALSE.

We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works

Both matrices must be the same size: TRUE

If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.

For any value of i, j; aij = bij: TRUE

Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.

8 0
3 years ago
Read 2 more answers
a car purchased for $34,000 is expected to lose value, or depreciate, at a rate of 6% per year.using x for years and y for the v
Vika [28.1K]

Answer:

y = 34000(1-0.06) ^ t

After 7.40 years it will be worth less than 21500

Step-by-step explanation:

This problem is solved using a compound interest function.

This function has the following formula:

y = P(1-n) ^ t

Where:

P is the initial price = $ 34,000

n is the depreciation rate = 0.06

t is the elapsed time

The equation that models this situation is:

y = 34000(1-0.06) ^ t

Now we want to know after how many years the car is worth less than $ 21500.

Then we do y = $ 21,500. and we clear t.

21500 = 34000(1-0.06) ^ t\\\\log(21500/34000) = tlog(1-0.06)\\\\t = \frac{log(21500/34000)}{log(1-0.06)}\\\\t = 7.40\ years.

After 7.40 years it will be worth less than 21500

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(k-7) would be the equation

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