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3 years ago

Define the value of the expression 3^2 x 3^{-5}

Mathematics

1 answer:

Tanya [424]3 years ago

8 0

Answer:

Between 0 and 1

Step-by-step explanation:

The given expresion is

${3}^{2} \times {3}^{ - 5}$

Recall that:

${a}^{m} \times {a}^{n} = {a}^{m + n}$

We apply this property of exponents to get;

${3}^{2} \times {3}^{ - 5} = {3}^{2 + - 5} = {3}^{ - 3}$

We rewrite to obtain a positive index as:

${3}^{2} \times {3}^{ - 5} = \frac{1}{ {3}^{3} } = \frac{1}{27}$

Therefore the value of the expression is between 0 and 1.

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Y"-3xy=0 given ordinary point x=0. power series

liberstina [14]

$\displaystyle y=\sum_{n\ge0}a_nx^n$
$\implies\displaystyle y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}$

$y''-3xy=0$
$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0$
$\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0$
$\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge3}a_{n-3}x^{n-2}=0$
$\displaystyle2a_2+\sum_{n\ge3}\bigg(n(n-1)a_n-3a_{n-3}\bigg)x^{n-2}=0$

This generates the recurrence relation

$\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_{n-3}=0&\text{for }n\ge3\end{cases}$

Because you have

$n(n-1)a_n-3a_{n-3}=0\implies a_n=\dfrac3{n(n-1)}a_{n-3}$

it follows that $a_2=0\implies a_5=a_8=a_{11}=\cdots=a_{n=3k-1}=0$ for all $k\ge1$ .

For $n=1,4,7,10,\ldots$ , you have

$a_1=a_1$
$a_4=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1$
$a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}$
$a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1$

so that, in general, for $n=3k-2$ , $k\ge1$ , you have

$a_{n=3k-2}=\dfrac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}a_1$

Now, for $n=0,3,6,9,\ldots$ , you have

$a_0=a_0$
$a_3=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0$
$a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0$
$a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0$

and so on, with a general pattern for $n=3k$ , $k\ge0$ , of

$a_{n=3k}=\dfrac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}a_0$

Putting everything together, we arrive at the solution

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_0\underbrace{\displaystyle\sum_{k\ge0}\frac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}x^{3k}}_{n=0,3,6,9,\ldots}+a_1\underbrace{\displaystyle\sum_{k\ge1}\frac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}x^{3k-2}}_{n=1,4,7,10,\ldots}$

To show this solution is sufficient, I've attached is a plot of the solution taking $y(0)=a_0=1$ and $y'(0)=a_1=0$ , with $n=6$ . (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)

6 0

3 years ago

The probability of rolling a 6 with a six-sided number cube is 1/6 Choose the

Pavel [41]

Answer:

UNLIKELY

Step-by-step explanation:

Given that:

Probability of rolling a six with a six sided nuner cube = 1/6

Required outcome / Total possible outcomes

Number of 6 / total sides of number cube = 1/6

The probability value obtained shows that there is a chance of getting a six ; however it is possible at about once in every 6 rolls of the dice.

Hence getting a six isn't impossible but the chance is small. A probability of 1/6 is the smallest that can be obtained for any of each number on the dice. Hence, it could be said that the likelihood is 'UNLIKELY'

7 0

3 years ago

10 m<br> 5.7 m<br> what is the area?

dalvyx [7]

Answer:

If it's a triangle, you just take that answer and multiply it by 0.5

Step-by-step explanation:

Well if you take the two numbers and multiply them together- that is what you get. When it comes to area, you take the two side lengths and multiply them together

5 0

2 years ago

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How much should they be charged?

deff fn [24]

About 500 dollars your welcome

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3 years ago

Which of the following is not a quadratic function

Pepsi [2]

Answer:

option B

Step-by-step explanation:

because quadratic equation means x^2+ or -ax+ or-b

3 0

3 years ago