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3 years ago

Simplify (This is a fraction) : 9^6 over 9^7

Mathematics

1 answer:

trapecia [35]3 years ago

3 0

when 2 exponents of like bases are divided, you can subtract the exponents

1. $\frac{9^6}{9^7} = \frac{1}{9}$

2. $\frac{9^2}{9^7} = \frac{1}{9^5}$

3. $9^5 = 9^5$

4. $\frac{1}{9^9} = \frac{1}{9^9}$

5. negative exponents mean move to opposite sides, so

$\frac{1}{9^{-5}} = \frac{9^5}{1}=9^5$

6. $\frac{1}{9^5} = \frac{1}{9^5}$

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What is the slope????

murzikaleks [220]

Answer:

-6

Step-by-step explanation:

is that quizzes join lol

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3 years ago

The length of the diogonals of a rectangular garden is 34m. If it's longer side measures 30m, find the perimeter of garden.

MakcuM [25]

Answer:

92 m

Step-by-step explanation:

Use the Pythagorean theorem to find the length of the shorter side.

Let w = length of the shorter side.

$w^{2} + 30^{2} = 34^{2} \\ w^{2} + 900 = 1156\\w^{2} = 256\\w = \sqrt{256} = 16$

The perimeter is 2l + 2w = 2(30) + 2(16) = 60 + 32 = 92 m

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3 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

5 0

3 years ago

Would you like Brainliest?

IceJOKER [234]

Answer: yes please and thank you in advance :)

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2 years ago

What is the confidence interval estimate of the population mean percentage change in the price per share of stock during the fir

I am Lyosha [343]

Answer:

(-0.1059 ; - 0.0337)

Step-by-step explanation:

The data table is attached in the picture below:

These is a matched pair design ; which requires taking the difference of the two values for each sample :

The mean and standard deviation of the difference will be used to construct the confidence interval :

The mean of difference, dbar = Σx/n = - 0.0698

The standard deviation of difference, Sd ;

Sd = [√Σ(d - dbar)²/(n-1)] = 0.1054

n = sample size = 25

The confidence interval :

dbar ± [TCritical * Sd/√n]

Tcritical at 90% ; df = n -1 = 25 -1

Tcritical(90% , 24) = 1.1711

C.I = - 0.0698 ± (1.711 * 0.1054/√25)

C.I = - 0.0698 ± 0.0361

C.I = (-0.1059 ; - 0.0337)

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3 years ago