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4 years ago

Is is 6 2/7 rational or irrational

Mathematics

1 answer:

steposvetlana [31]4 years ago

6 0

Step-by-step explanation:

6 2/7is rational

-19is rational but next I dont know

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(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

Is 8,469,576 divisible by 6?

Nataly [62]

Yes because the last digit ends as an even number and 6 is also even, implying that it can be divisible by 6.

The answer would be 1,411,596 which of course is even.

Hope this helps.

6 0

3 years ago

Read 2 more answers

Find the area please! *you can use decimals if needed*

Lorico [155]

Answer:

116 km squared

Step-by-step explanation:

When you get crazy shapes like this, try and look for the basic ones inside (i.e. triangles, rectangles, etc.) I can find one big triangle and two rectangles, one small and one larger. Vide the attatchment for reference.

Step 1. Find the missing variables, x, y, and z

y = 2

x = 6 - 2 = 4

z = 8 + (4) + 2 = 14

Step 2. Find the area of each shape

Triangle:

Area = bh/2 = (8 x 14)/2 = 56 km sq

Big rectangle:

Area = bh = 8 x 6 = 48 km sq

Small rectangle:

Area = bh = 2 x 6 = 12 km sq

Step 3. Add all the areas together

56 + 48 + 12 = 116 km sq

I hope this helps!

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5 0

2 years ago

Read 2 more answers

Which number is irrational? a. .33333333 b. .03030303 c. .993993993 d. .131131113...

Effectus [21]

Repeating decimals are rational....but they have to be repeating.
The only one that is not repeating is : D .131131113...now if the last 3 digits would have been 131 instead of 113, then it would have been rational.

8 0

3 years ago

Help Please! would really appreciate it.

muminat

<h3>Answer: x = 9</h3>

=========================================

The diagram shows:

base = 13
height = x

Recall that the area of a parallelogram is equal to the base times height

area = base*height

We're told that the area is 117 square units, so that means 13x = 117.

Divide both sides by 13 to isolate x

13x = 117

13x/13 = 117/13

x = 9

The height of the parallelogram is 9

Note how:

area = base*height = 13*x = 13*9 = 117

which helps confirm we have the correct height value for x.

8 0

3 years ago