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3 years ago

Find the measure of Angle A A.) 80 B.) 40 C.) 60 D.) 20

Mathematics

1 answer:

olasank [31]3 years ago

7 0

I would assume either 20 or 40 but i rly think it’s 20. sorry if i’m wrong

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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

Can someone give me the slope and y-intercept for this problem

aivan3 [116]

Answer:

Slope: 2

Y-intercept: -4

Step-by-step explanation:

Slope = y2 - y1 / x2 - x1

Slope = -2 - -4 / 1 - 0

Slope = 2 / 1

Slope = 2

Y-intercept is when x = 0

This is shown on the first row

When x = 0, y = -4, which is our y-intercept

3 0

3 years ago

Solve for x. 5x - 4 = -3x+ 12 X = 2 X=6 X=-4

marusya05 [52]

Answer:

x = 2

Step-by-step explanation:

Given

5x - 4 = - 3x + 12 ( add 3x to both sides )

8x - 4 = 12 ( add 4 to both sides )

8x = 16 ( divide both sides by 8 )

x = 2

5 0

3 years ago

−(1/4w+8)+8=3 the 1/4 is a fraction

Dvinal [7]

Answer:

w = -12

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How do I solve tan35=5.5/5.5+x ?

nlexa [21]

Answer:

Solve for

by simplifying both sides of the equation, then isolating the variable.

Step-by-step explanation:

is this solve for x? you can just tell me what you are doing with this ex: adding, etc

3 0

3 years ago