geduiwelh
2021-02-05
Answered

Solve the linear congruence

$x+2y\equiv 1\left(\text{mod}5\right)$

$2x+y\equiv 1\left(\text{mod}5\right)$

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Latisha Oneil

Answered 2021-02-06
Author has **100** answers

Step 1

Given a system of linear congruences

$x+2y\equiv 1\left(\text{mod}5\right)$

$2x+y\equiv 1\left(\text{mod}5\right)$

Solve it.

Step 2

Multiply the first congruence by 2.

$2x+4y\equiv 2\left(\text{mod}5\right)$

Add the preceding one with the second congruence in the system.

$4x+5y\equiv 3\left(\text{mod}5\right)$

$\Rightarrow 4x\equiv 3\left(\text{mod}5\right)$

Step 3

Multiply the congruence by${4}^{-1}\left(\text{mod}5\right)=4$ to get

$16x\equiv 12\left(\text{mod}5\right)$

$\Rightarrow x\equiv 2\left(\text{mod}5\right)$

Step 4

Plugging back in x for the second equation.

$4+y\equiv 1\left(\text{mod}5\right)$

$\Rightarrow y\equiv -3\left(\text{mod}5\right){\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}y\equiv 2\left(\text{mod}5\right)$

Thus the solution of the system of congruences is

$x\equiv 2\left(\text{mod}5\right),y\equiv 2\left(\text{mod}5\right)$

Given a system of linear congruences

Solve it.

Step 2

Multiply the first congruence by 2.

Add the preceding one with the second congruence in the system.

Step 3

Multiply the congruence by

Step 4

Plugging back in x for the second equation.

Thus the solution of the system of congruences is

asked 2022-07-16

A different approach to a common question

A unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?

This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is $\frac{5}{18}$, we reach this conclusion by using $E(L+M+S)=1$, and we know we can calculate E(L) and E(S), so we just use $E(M)=1-E(L)-E(S)$ to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?

L: Length of the longest part

S: Length of the smallest part

M: Length of the medium part

A unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?

This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is $\frac{5}{18}$, we reach this conclusion by using $E(L+M+S)=1$, and we know we can calculate E(L) and E(S), so we just use $E(M)=1-E(L)-E(S)$ to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?

L: Length of the longest part

S: Length of the smallest part

M: Length of the medium part

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Euclidean norm vs pythagorean theorem?

I recently stumbled upon an Euclidian norm. First I thought there are the powers and square root to deal with possible negative values (like in Standard deviation formula) but then I realized, the final number (sum of squares and square root of it) is not same as sum of absolute numbers.

So then I noticed it is more related to Pythagorean theorem (Euclidian distance).

But if looked in google for PT in 3d space i found the formula like

$({a}^{2}+{b}^{2}+{c}^{2}{)}^{\frac{1}{2}}$

But in the Euclidian norm it is

$SQRT({a}^{2}+{b}^{2}+{c}^{2})$

How comes, it is different?

Second question, how comes that the theorem works for any count of dimensions?

I recently stumbled upon an Euclidian norm. First I thought there are the powers and square root to deal with possible negative values (like in Standard deviation formula) but then I realized, the final number (sum of squares and square root of it) is not same as sum of absolute numbers.

So then I noticed it is more related to Pythagorean theorem (Euclidian distance).

But if looked in google for PT in 3d space i found the formula like

$({a}^{2}+{b}^{2}+{c}^{2}{)}^{\frac{1}{2}}$

But in the Euclidian norm it is

$SQRT({a}^{2}+{b}^{2}+{c}^{2})$

How comes, it is different?

Second question, how comes that the theorem works for any count of dimensions?

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Given a right circular cone with the line of symmetry along $x=0$, and the base along $y=0$, how can I find the maximum volume paraboloid (parabola revolved around the y-axis) inscribed within the cone? Maximising the volume of the paraboloid relative to the volume of the right circular cone. In 2-D, the parabola has 2 points of tangency to the triangle, one of each side of the line of symmetry. I have tried using the disk method to find the volume of the cone, and the parabola, both with arbitrary equations such as $y=b-ax$, and $y=c-d{x}^{2}$, but I end up with a massive equation for several variables, instead of a simple percentage answer. Any help is appreciated! Thanks in advance.

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A coffee company sells two types of breakfast blends. They have on hand 132 kg of dark roast and 84 kg of hazelnut. One breakfast blend will contain half dark roast and half hazelnut and will sell for $\mathrm{\$}7$ per kg. The other breakfast blend will contain 3/4 dark roast and 1/4 hazelnut and will sell for $\mathrm{\$}9.50$ per kg.

My approach:

I see this is an optimization problem and need to show that $132{x}_{1}+84{x}_{2}$ needs to be optimized but confused on the other constraints

I used the convex optimization polygon theory of checking on the endpoints and computing the maximal value, can someone check and elaborate further on the reply below?

My approach:

I see this is an optimization problem and need to show that $132{x}_{1}+84{x}_{2}$ needs to be optimized but confused on the other constraints

I used the convex optimization polygon theory of checking on the endpoints and computing the maximal value, can someone check and elaborate further on the reply below?

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