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Gemiola [76]
3 years ago
11

There is a language with 12 distinct letters. How many words that are less than four letters long can possibly be formed? (Assum

e any combination of letters is valid and letters can be repeated). Can you help me out with this one?
Mathematics
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

1884

Step-by-step explanation:

1)If you hear the condition "less than four letters" it means that each word can consist of 3,2 or 1 letter. Firstly, consider the easiest situation with the word from one letter. There are 12 letters, so there are 12 words from one letter.

Then the words from two letters. The first letter of such a word can be chosen in 12 means(because 12 letters are available), then the second letter can be chosen in 12 means too. 12*12=144 words with 2 letters. (Multiplying, not adding, because each two words form one combination).

Then for the words from three letters, use the same rule, 12 means for the first letter, 12 means for the second letter, 12 means for the third letter. 12*12*12=1728 words.

Having particular words from 1,2,3 letters add them to get total quantity of words: 12+144+1728=1884words,that is the answer.

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A 500 cm piece of wire with a radius of 0.2 cm has a resistance of 0.025 ohm. Find an equation that relates these variables.
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Answer:

Step-by-step explanation:

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3 years ago
y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the
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The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.

If we add the equations above, we get:

\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,

which do correspond to the desired initial conditions.

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3 years ago
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or if you know the derivative of \tan:

\cot x=\dfrac1{\tan x}

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or if you don't know the derivative of \csc,

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which is the same as the previous result since

\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}

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Answer:

A.–1.8

Step-by-step explanation:

5 0
3 years ago
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