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3 years ago

6

A combination lock requires 4 selections of numbers, each from 1 to 20. • how many different combinations are there? • suppose t

hat the lock is constructed in such a way that no number maybe used twice. how many different combinations are possible?

1 answer:

slamgirl [31]3 years ago

6 0

NUmber of combination = 20 x 19 x 18 x 17 = 116 280

Answer: 116 280

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If point p and q contened in a plane then PQ is entirely contained in that plane?

anyanavicka [17]

Please elaborate on this so that I can help you.

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3 years ago

In Omak the sales tax is 8.4%. What is the total cost including tax Aakel spent on a $48 purchase?

Vikki [24]

Answer:

$52.03

Step-by-step explanation:

1. Approach

To solve this problem, first one needs to calculate the sales tax, then one must add that to the amount spent on the purchase. To calculate sales tax, one must convert the percent to decimal form, this can be done by dividing the percent by 100. Then one will multiply the decimal by the amount spent.

2. Find the tax

As states above; to calculate sales tax, one must convert the percent to decimal form, this can be done by dividing the percent by 100. Then one will multiply the decimal by the amount spent.

<u>a. convert percent to decimal</u>

8.4 / 100 = 0.084

<u>b. multiply the decimal by the amount spent</u>

48 * 0.084 = 4.032

The amount spent on sales tax is, $4.032

3. Find the total amount spent

Now all one has to do is add the amount spent in tax by the amount spent on the purchase.

48 + 4.032 = 52.032

Since money is only spent rounded to the second decimal point, one has to round the number;

52.03

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2 years ago

Solve x √y=21 y √x=29

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3 years ago

The ratio of the number of students to the number of teachers in a school is 17:5. If there are 35 teachers in a school, how man

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Step-by-step explanation:

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago