Khadija Wells
2021-03-06
Answered

Solve the following systems of congruences.

$x\equiv 4(mod\text{}7)$

You can still ask an expert for help

Jaylen Fountain

Answered 2021-03-07
Author has **170** answers

Formula used:

1) Theorem: System of congruences:

Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences

Furthermore, any two solutions x and y are congruent modulo mn.

2) Theorem: Addition and Multiplication Properties:

If

3) Theorem: Cancellation Law:

If

Explanation:

Consider

By using addition property,

By using multiplication property,

Since

Therefore, the system of congruences is

Since 7 and 8 are relatively prime then

Then, by using theorem there exists an integer x that satisfies the system of congruences.

From the first congruence

By using addition property,

Since

Since (7, 8) = I then by using cancellation law,

Thus,

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Pythagorean theorem in higher dimensions?

When using vector notation in coordinate systems (Cartesian coordinates) we see that the magnitude of a vector in two dimensions is equal to the square root of its Y component squared added to its X component squared (Pythagorean theorem).But the same calculation is done for a three dimensional vector that has X, Y, and Z components.Is there a triangle that has four sides? (of course not, but how does this right triangle formula work for a calculation that involves more than two dimensions?).

When using vector notation in coordinate systems (Cartesian coordinates) we see that the magnitude of a vector in two dimensions is equal to the square root of its Y component squared added to its X component squared (Pythagorean theorem).But the same calculation is done for a three dimensional vector that has X, Y, and Z components.Is there a triangle that has four sides? (of course not, but how does this right triangle formula work for a calculation that involves more than two dimensions?).

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Sketch the following conic sections:

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b)${y}^{2}-4y-4{x}^{2}=8$

c)$3{(y-5)}^{2}-7{(x+1)}^{2}=1$

a)

b)

c)

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To prove: In the following figure, triangles ABC and ADC are congruent.

Given:

Figure is shown below:

It is provided that$\mathrm{\angle}BAC=\mathrm{\angle}DAC\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\mathrm{\angle}BCA=\mathrm{\angle}DCA$ .

Given:

Figure is shown below:

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Finding the maximum volume of a cylinder.

The cylinder is open topped and is made from $200\text{}{\text{cm}}^{2}$ of material.

So what I have done is said $S=2\pi {r}^{2}h+2\pi {r}^{2}$.

$\begin{array}{rl}2\pi {r}^{2}h+2\pi {r}^{2}& =200\\ 2\pi {r}^{2}h& =200-2\pi {r}^{2}\\ h& =\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}\end{array}$

Subbing this into the volume equation I get $V=(\pi {r}^{2})\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}$.

Having trouble differentiating that equation.

The cylinder is open topped and is made from $200\text{}{\text{cm}}^{2}$ of material.

So what I have done is said $S=2\pi {r}^{2}h+2\pi {r}^{2}$.

$\begin{array}{rl}2\pi {r}^{2}h+2\pi {r}^{2}& =200\\ 2\pi {r}^{2}h& =200-2\pi {r}^{2}\\ h& =\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}\end{array}$

Subbing this into the volume equation I get $V=(\pi {r}^{2})\frac{200-2\pi {r}^{2}}{2\pi {r}^{2}}$.

Having trouble differentiating that equation.

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Distance between (5,-11,-5) and (4,-13,6)?