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castortr0y [4]
3 years ago
9

Choose the equation that represents the graph below.

Mathematics
2 answers:
joja [24]3 years ago
7 0

Answer:

c. y=-\frac{2}{3}x+6

Step-by-step explanation:

We have to choose that equation which represents the given graph.

From given graph , the graph cut the y- axis at point (0,6) and cut the x- axis (9,0).

a.y=-2x-6

Substitute x=0 then we get

y=-6

But we have y-intercept y=6

Therefore, this equation does not represents the given graph.

b. y=\frac{2}{3}x+6

Substitute  y=0 then, we get

\frac{2}{3}x+6=0

\frac{2}{3}x=-6

x=\frac{-6\times 3}{2}=-9

The x- intercept =-9

But , we have x=9

Therefore, it is false.

c. y=-\frac{2}{3}x+6

Substitute x=0 then we get

y=6

Substitute y=0

Then, we get

-\frac{2}{3}x+6=0

\frac{2}{3}x=6

x=\frac{6\times 3}{2}=9

Hence, the line passing through the point (9,0) and (0,6) .Therefore, it represents the given graph.

Answer:c. y=-\frac{2}{3}x+6

alekssr [168]3 years ago
5 0
The equation is y = -2/3x + 6
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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

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

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

x=2\cdot7+5\cdot6

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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

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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}.

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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}}

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Answer:

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Step-by-step explanation:

HK and JL are diagonals of the parallelogram.

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