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3 years ago

3. Complete the proof below. Help please!!!

Mathematics

1 answer:

baherus [9]3 years ago

5 0

Answer:

The completed proof is presented as follows;

The two column proof is presented as follows;

Statements ${}$ Reason

1. $\overline {HI}$ ║ $\overline {KL}$ , J is the midpoint of $\overline {HL}$ ${}$ 1. Given

2. ∠IHJ ≅ ∠JLK ${}$ 2. Alternate angles are congruent

3. ∠IJH ≅ ∠KJL ${}$ 3. Vertically opposite angles

4. $\overline {HJ}$ ≅ $\overline {JL}$ ${}$ 4. Definition of midpoint

5. ΔHIJ ≅ ΔLKJ ${}$ 5. By ASA rule of congruency

Step-by-step explanation:

Alternate angles formed by the crossing of the two parallel lines $\overline {HI}$ and $\overline {KL}$ , by the transversal $\overline {HL}$ are equal

Vertically opposite angles formed by the crossing of two straight lines $\overline {IK}$ and $\overline {HL}$ are always equal

A midpoint divides a line into two equal halves

Angle-Side-Angle, ASA rule of congruency states that two triangles ΔHIJ and ΔLKJ, that have two congruent angles, ∠IHJ in ΔHIJ ≅ ∠JLK ${}$ in ΔLKJ and ∠IJH in ΔHIJ ≅ ∠KJL in ΔLKJ, and that the included sides between the two congruent angles is also congruent $\overline {HJ}$ ≅ $\overline {JL}$ , then the two triangles are congruent, ΔHIJ ≅ ΔLKJ.

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3 years ago

Find an integer x such that 0<=x<527 and x^37===3 mod 527

Greeley [361]

Since $527=17\times31$ , we have that

$x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}$

By Fermat's little theorem, and the fact that $37=2(17)+3=1(31)+6$ , we know that

$x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}$
$x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}$

so we have

$\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}$

Consider the first case. By Fermat's little theorem, we know that

$x^{17}\equiv x^{16}x\equiv x\mod{17}$

so if we were to raise $x^5$ to the $n$ th power such that

$(x^5)^n\equiv x^{5n}\equiv x\mod{17}$

we would need to choose $n$ such that $5n\equiv1\mod{16}$ (because $16+1\equiv1\mod{16}$ ). We can find such an $n$ by applying the Euclidean algorithm:

$16=3(5)+1$
$\implies1=16-3(5)$
$\implies16-3(5)\equiv-3(5)\equiv1\mod{16}$

which makes $-3\equiv13\mod{16}$ the inverse of 5 modulo 16, and so $n=13$ .

Now,

$x^5\equiv3\mod{17}$
$\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}$

$3^1\equiv3\mod{17}$
$3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}$
$3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}$
$\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}$

Similarly, we can look for $m$ such that $7m\equiv1\mod{30}$ . Apply the Euclidean algorithm:

$30=4(7)+2$
$7=3(2)+1$
$\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)$
$\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}$

so that $m=13$ is also the inverse of 7 modulo 30.

And similarly,

$x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}$

Decomposing the power of 3 in a similar fashion, we have

$3^{13}\equiv(3^3)^4\times3\mod{31}$

$3\equiv3\mod{31}$
$3^3\equiv27\equiv-4\mod{31}$
$\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}$

So we have two linear congruences,

$\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}$

and because $\mathrm{gcd}\,(17,31)=1$ , we can use the Chinese remainder theorem to solve for $x$ .

Suppose $x=31+17$ . Then modulo 17, we have

$x\equiv31\equiv14\mod{17}$

but we want to obtain $x\equiv12\mod{17}$ . So let's assume $x=31y+17$ , so that modulo 17 this reduces to

$x\equiv31y+17\equiv14y\equiv1\mod{17}$

Using the Euclidean algorithm:

$17=1(14)+3$
$14=4(3)+2$
$3=1(2)+1$
$\implies1=3-2=5(3)-14=5(17)-6(14)$
$\implies-6(14)\equiv11(14)\equiv1\mod{17}$

we find that $y=11$ is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

$x\equiv31(11)(12)+17\equiv12\mod{17}$

To satisfy the second condition that $x\equiv24\mod{31}$ , taking $x$ modulo 31 gives

$x\equiv31(11)(12)+17\equiv17\mod{31}$

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find $z$ such that $17z\equiv1\mod{31}$ . Euclidean algorithm:

$31=1(17)+14$
$17=1(14)+3$

and so on - we've already done this. So $z=11$ is the inverse of 17 modulo 31. Now, we take

$x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}$

as required. This means the congruence $x^{37}\equiv3\mod{527}$ is satisfied by

$x=31(11)(12)+17(11)(24)=8580$

We want $0\le x$ , so just subtract as many multples of 527 from 8580 until this occurs.

$8580=16(527)+148\implies x=148$

3 0

3 years ago

Help i dont how to do this

IrinaK [193]

What is it you are trying to do I can help if you just say it.

4 0

3 years ago

In 2010 the population of alaska was about 710200.What ir this written in Word form

Mazyrski [523]

Answer:

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Step-by-step explanation:

6 digits would mean it's 100k, so you just go from there. Write out the numbers in word form adding "hundred" and "thousand" as you must. This applies to all problems similar to this.

7 0

3 years ago