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3 years ago

Passes through (3,-6) and (-1, 2)

Mathematics

1 answer:

hram777 [196]3 years ago

7 0

Answer:

Therefore, Equation of line and the slope is

$y=-2x$

$Slope = -2$

Step-by-step explanation:

Given:

Let,

point A( x₁ , y₁) ≡ ( 3 ,-6)

point B( x₂ , y₂ )≡ (-1 , 2)

To Find:

Equation of Line AB =? ( Assumption)

Solution:

Equation of a line passing through a points A( x₁ , y₁) and point B( x₂ , y₂ ) is given by the formula Two -Point Form,

$(y-y_{1})=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }\times (x-x_{1})$

Now on substituting the slope and point A( x₁ , y₁) ≡ ( 3 , 6) and B( x₂ , y₂ )≡ (-1 , 2) we get

$(y-(-6))=\dfrac{2--6}{-1-3}\times (x-3)\\\\y+6=\dfrac{8}{-4}(x-3)\\\\y+6=-2(x-3)\\y+6=-2x+6\\y=-2x$ .....Required Equation of line

Which is also in the form of

$y=mx$ ....Also called as Equation of line Passing through Origin.

where , m =slope

Therefore, Equation of line and the slope is

$y=-2x$

$Slope = -2$

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

tino4ka555 [31]

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

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.

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

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

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

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 $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

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3 years ago

a man bought a motor at 400,000 Kenyan shillings in January 1999 it depreciated at a rate of 16% per annum if he value it six mo

Alexxx [7]

The value of the car is January 2003 is $199,148.54.

<h3>What is the value of the car?</h3>

Depreciation is the rate of decline in the value of an asset with the passage of time.

The exponential equation that can be used to determine the value of the car is:

Value of the car = purchase value(1 - rate of decline)^time

400,000 x (1 - 0.16)^(2003 - 1999)

400,000 x (0.84^4) = $199,148.54

To learn more about depreciation, please check: brainly.com/question/15085226

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Which sequence can be generated from the formula f(x + 1) = One-half(f(x))?

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Step-by-step explanation:

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