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Which expression is equivalent to [(3xy^-5)^3 / (x^-2y^2)^-4]^-2

True [87]

The answer is (x¹⁰y¹⁴)/729.

Explanation:
We can begin simplifying inside the innermost parentheses using the properties of exponents. The power of a power property says when you raise a power to a power, you multiply the exponents. This gives us

[(3³x³y⁻¹⁵)/(x⁸y⁻⁸)]⁻².

Negative exponents tell us to "flip" sides of the fraction, so within the parentheses we have
[(3³x³y⁸)/(x⁸y¹⁵)]⁻².

Using the quotient property, we subtract exponents when dividing powers, which gives us
(3³/x⁵y⁷)⁻².

Evaluating 3³, we have
(27/x⁵y⁷)⁻².

Using the power of a power property again, we have
27⁻²/x⁻¹⁰y⁻¹⁴.

Flipping the negative exponents again gives us x¹⁰y¹⁴/729.

4 0

3 years ago

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PLEASE HELP!!!!

stira [4]

"The total number of stamps is 35" is the statement among the following choices given in the question that is correct. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that has actually come to your great help.

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4 years ago

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is