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3 years ago

Guys help me out please and thanks :)

Mathematics

1 answer:

Jobisdone [24]3 years ago

5 0

The answer is D. BC

To find an opposite of an angle, it will always be a line, not an angle.
The line that is directly opposite from the angle A is line BC.

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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

I inserted a picture so it can be more clear. This is for Algebra 2, Unit 2

dedylja [7]

If f(x) = 3x-4 and g(x) = 4-3x then the solution of f(g(x)) = g(f(x)) proves that they are not inverse functions.

According to the question,

We have the following two functions:

f(x) = 3x-4 and g(x) = 4-3x

We know that in order to find whether two functions are inverse of each other or not we need to find the value of f(g(x)) and g(f(x)) and then see whether they are equal to x.

Now, among the given options, the only option representing this kind of solution is option D and the solutions are not equal which proves that they are not inverse functions.

Hence, the correct option is D.

To know more about inverse functions here

brainly.com/question/2541698

#SPJ1

4 0

1 year ago

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!

jenyasd209 [6]

Answer:

x= -1/3

y= -3 1/3

Step-by-step explanation:

Make both of them into y=Mx+b form

y=-x-3

y=(-x-7)/2

Substitute the x-3 in the first equation for the y in the second one.

x-3=(-x-7)/2

2x-6=-x-7

3x=-1

x=-1/3

Now that we have x, we can solve for y.

y=-x-3

y=-1/3-3

y=-3 1/3

7 0

3 years ago

Helppppp meeeeee pleaseeeeeee

Darina [25.2K]

Correct answer is B……..

3 0

3 years ago

Read 2 more answers

Factor 140c +28- 14a

saul85 [17]

The answer would be C if you factored out a 14.

5 0

3 years ago

Read 2 more answers