All categories

3 years ago

Please help, i'll give brainliest

Mathematics

1 answer:

qaws [65]3 years ago

3 0

Answer:

Step-by-step explanation:

You want ...

4x +24 = 8x

24 = 4x . . . . . . . . subtract 4x

6 = x . . . . . . . . . . .divide by 4

When x=6, the slats are parallel.

You might be interested in

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

5 0

2 years ago

Pls help me answer 2.2.3 and 2.2.4 (2.2.4 needs to be converted) ill give brainstorm

Anarel [89]

$\bold{\huge{\underline{ Solution }}}$

<h3>Analysis of graph :-</h3>

We have given one graph which is plot between distance and time. where time is in minutes and distance is in seconds.

According to the graph

For the first 5 minute ( O to A) , The distance is continously increasing 2m / per minute .
For the 5 minute that is from 5 minute to 13 minute ( A to B) both marellize and her dog wally moving with the constant speed .
For next 3 minutes that is from 10 minutes to 13 minutes ( B to C) , The distance is continously decreasing with time .
For next 3 minutes that is from 13 to 16 minutes ( C to D) , Again they moved with constant speed .
For next 6 minute that is from 16 to 21 minutes ( D to E) . Again, There distance is increasing with time .
Again For next 4 minutes that is 21 to 25 minutes , they are moving with constant velocity .

<h3>Let's Begin :-</h3>

1) Between O and A

The marellize and wally when moving between O to A , The distance is constantly increasing with time.
The graph is Straight line

2) Between A and B

The marellize and wally when moving between A to B, The distance remains the same with time that is they moving with constant speed.
The graph is constant or steady

3) Between B to C

The marellize and wally when moving between B to C, The distance is constantly decreasing with time .
The graph is straight line but it follows decreasing function .

4) For covering the First 6 km ,

According to the graph,