Ask question

Ask question

All categories

3 years ago

6

I have $5$ different mathematics textbooks and $4$ different psychology textbooks. In how many ways can I place the $9$ textbook

s on a bookshelf, in a row, if there must be a psychology textbook exactly in the middle, and there must be a mathematics textbook at each end?

1 answer:

leonid [27]3 years ago

6 0

Answer:

What

Step-by-step explanation:

You can’t have more than 2 combinations with £9

You might be interested in

(a) the number 561 factors as 3 · 11 · 17. first use fermat's little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11),

Vitek1552 [10]

LFT says that for any prime modulus $p$ and any integer $n$ , we have

$n^p\equiv n\pmod p$

From this we immediately know that

$a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}$

Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case $11\times17=187=62\times3+1$ , so

$a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3$

Next, $63=21\times3$ , so

$a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3$

Next, $21=7\times3$ , so

$a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3$

Finally, $7=2\times3+1$ , so

$a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3$

We do the same thing for the remaining two cases:

$3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}$

$3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}$

Now recall the Chinese remainder theorem, which says if $x\equiv a\pmod n$ and $x\equiv b\pmod m$ , with $m,n$ relatively prime, then $x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}$ , where ${m_n}^{-1}$ denotes $m^{-1}\pmod n$ .

For this problem, the CRT is saying that, since $a^{561}\equiv a\pmod3$ and $a^{561}\equiv a\pmod{11}$ , it follows that

$a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}$
$\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}$
$\implies a^{561}\equiv 34a\equiv a\pmod{33}$

And since $a^{561}\equiv a\pmod{17}$ , we also have

$a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}$
$\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}$
$\implies a^{561}\equiv562a\equiv a\pmod{561}$

6 0

4 years ago

Kate pays $36 for 2 sets of candles. Each set contains 3 red candles and 3 white candles. What is the cost per candle in these s

BabaBlast [244]

Answer: A

Step-by-step explanation: Add the amount of candles for both sets together (6 in each set, so 12) and then divide 36 by 12 for 3.

8 0

4 years ago

Read 2 more answers

Which steps transform the graph of y=x^2 to y=-(x-3)^2+2

damaskus [11]

Answer:

reflection over the x-axis

shifted 3 right

shifted 2 up

Step-by-step explanation:

This function is quadratic. Quadratic graphs are written as y=a(x-h)^2+k. Where a is a non-zero number that affects the vertical appearance of the graph, h is the horizontal placement, and k is the vertical placement.

When a is negative then, the graph has been reflected over the x-axis. This makes it look like it has been turned upside down. The variable h affects how the x-values are shifted. In this equation, h looks negative but it is actually positive because if you look at the formula, h is being subtracted. Positive h values move the graph right. So, the graph is shifted 3 units to the right. Finally, k does the same thing as h but on the vertical (y) axis. So, +2 makes the graph shift 2 units up.

8 0

2 years ago

You have $124. You can earn $22 each time you mow your neighbor's lawn. The function s(x)=22x+124 represents your total savings.

hodyreva [135]

S(x) = 22x + 124 s(5) = 22(5) + 124 s(5) = $234 300 = 22x + 124 -124 -124 176 = 22x divide by 22 8 = x

7 0

3 years ago

Read 2 more answers

A regular decagon has 10 sides. How many reflectional symmetries does a regular decagon have?

lina2011 [118]

The rule for regular polygons are very easy. The number of reflectional symmetries is same as the number of sides. Regular polygons have all sides the same length and all angles same. Reflection symmetry means that you can fold the shape along that line and it will match up.

For this question, we want the number of reflectional symmetries of a regular decagon. Decagon is a 10 sided figure. Hence, the number of reflectional symmetries is 10.

There are 5 symmetry lines from one vertex to opposite vertex and 5 more symmetry lines form midpoint of one side to midpoint of opposite side.

ANSWER: 10

7 0

4 years ago

Read 2 more answers