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3 years ago

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Which of the following is an example of a quadratic equation?

2 answers:

Gre4nikov [31]3 years ago

6 0

The answer should be B, i hope this helps

REY [17]3 years ago

3 0

Answer:

B. x^2 - 64 = 0

Step-by-step explanation:

A. x+10 = 33

B. x^2 - 64 = 0

C. Y-776

D. y = 4x + 6

A quadratic has an term with a power of two

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Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

a.

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

$x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7$

5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

$\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}$

b.

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

5 0

3 years ago

Suda çözündüklerinde H* iyonu verebilen

Afina-wow [57]

Answer:

u just need to pute it right

7 0

3 years ago

What is the meaning of the unknown factor and quotient

Komok [63]

I don't know what you mean by unknown factor but a quotient is the answer of a division problem

7 0

3 years ago

What is 2(5-3v)+9v=28?

almond37 [142]

10-6v+9v=28
10+3v=28
-10 -10
3v=18
3 3
v=6

7 0

3 years ago

Read 2 more answers

Frank invested $30,000 at 4% simple interest. How much interest will he earn each year?

Oduvanchick [21]

Answer:

$1200 interest

Step-by-step explanation:

$30,000 divided by 100 = 300 , 300 multipled by 4 = 1200 , thats how to find what 4% would be

3 0

3 years ago