GYanggg

\(x^3=125 \\ x^3=5^3 \\ x=5 \)  

I hope this helped,

Gavin

Thanks, I'n just glad I can help. 

Hi ant101!

We use the rule that states:

In the quadratic equation \(ax^2+bx+c=0\), 

The sum of the roots is \(-\frac{b}{a}\)

The produdct of the roots is \(\frac{c}{a}\)

In your question, the sum of the two roots is \(2-3i+3i+2=4\)

The product is, \((2-3i)(2+3i)\)

Using \(a^2-b^2=(a-b)(a+b)\), 

\((2-3i)(2+3i)=2^2-(3i)^2=4-(9\cdot(-1))=4+9=13\)

Using the law, \(-\frac{a}{1}=4 \\ \frac{b}{1}=13\)

\(a=-4 \\ b=13 \\ 13+(-4)=\boxed9\)

That is the answer you seek.

I hope this helped,

Gavin

3.

The modular multiplicative inverse of an integer a modulo m is an integer b such that

\(ab \equiv 1 \pmod m\), It maybe noted \(a^{-1}\), where the fact that the inversion is m-modular is implicit.

The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1).

If the modular multiplicative inverse of a modulo m exists, the operation of division by a modulo m can be defined as multiplying by the inverse.

Zero has no modular multiplicative inverse.

The modular multiplicative inverse of a modulo m can be found with the Extended Euclidean algorithm.

To show this, let's look at this equation: ax + my = 1

This is linear diophantine equation with two unknowns, refer to Linear Diophantine equations.

Since one can be divided without reminder only by one, this equation has the solution only if \( {\rm gcd}(a,m)=1\). The solution can be found with the Extended Euclidean algorithm. The modulo operation on both parts of equation gives us \(ax = 1 \pmod m \)Thus, x is the modular multiplicative inverse of a modulo m.

The answer you seek is 180

I hope this helped,

gavin

Desmos is correct.

\(|x+3|=2 \\\text{should not be a slanted line because both} \\(x+3) \text{and}-(x+3) \text{is equal to 2} \)

To solve:

\((x+3)=2 \\ x=-1\)

\(-(x+3)=2 \\ -x-3=2 \\ -x=5 \\ x=-5\)

Two vertical lines, x = -1, and x = -5 is the graph of your equation. 

Here is the full solution.

Just as guest said.

1. 

\(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=\frac{25-25+a}{25-a}+\frac{65-65+b}{65-b}+\frac{60-60+c}{60-c}\)

\((\frac{25}{25-a}-1)+(\frac{65}{65-1}-1)+(\frac{60}{60-c}-1)=\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}-3\)

After factoring out the five, we have:

\(5(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c})-3=7\)

We can see that the value of what we are trying to find and its relantionship to the original expression.

Since the original expression is 7, and the expression we are trying to find is 3 less than 5 times 7,

We have:

\((7+3)\div5=2\)

Just the way guest did it!

I hope this helped,

Gavin.

According to the laws of geometry, 

Angle B is one half of arc AC

\((4x-3.5)\cdot2=4x+17\)

\(8x-7=4x+17\)

\(4x=24\)

\(x=6\)

Plugging this into 4x + 17, we get:

\(4\cdot6+17=\boxed{41}\)

I hope this helped,

Gavin

Hi Guest!

We can rewrite the left side of the equation,

using the arithmetic series sum formula:

\(S_n=\frac{n(a_1+a_n)}{2}\)

We get:

\(100=\frac{n(2a+n-1)}{2}\)

\(200=n(2a+n-1)\)

Hence, n must be a factor of 200.

 \( \begin{array}{c|c|c} n & 2a + n - 1 & a \\ \hline 2 & 100 & 99/2 \\ 4 & 50 & 47/2 \\ 5 & 40 & 18 \\ 8 & 25 & 9 \\ 10 & 20 & 11/2 \\ 20 & 10 & -9/2 \\ 25 & 8 & -8 \\ 40 & 5 & -17 \\ 50 & 4 & -45/2 \\ 100 & 2 & -97/2 \\ 200 & 1 & -99 \end{array} \ \)

Since a must be a positive integer, the solutions are \((a,n) = \boxed{(18,5)\text{ and }(9,8)}\). 

I hope this helped,

Gavin

A box and whisker plot—also called a box plot—displays the five-number summary of a set of data.

The five-number summary is the minimum, first quartile, median, third quartile, and maximum.

In a box plot, we draw a box from the first quartile to the third quartile.

Hey Maplesnowy!

Since a full circle is 360 degrees and we are missing one angle, we can write an equation to solve for that angle.

Also, a right angle is 90 degrees.

90 + 90 + 150 + x = 360

330 + x = 360

x = 30

The angle you want to solve for is 30 degrees.

I hope this helped,

Gavin.