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\(x^3=125 \\ x^3=5^3 \\ x=5 \)  

I hope this helped,

Gavin

2. What is the minimum value of the expression 

2x^2+3y^2+8x-24y+62 for real  \(x\)  and  \(y\)?

\(\begin{array}{|rcll|} \hline && 2x^2+3y^2+8x-24y+62 \\ &=& 2x^2+8x +3y^2-24y+62 \\ &=& 2(x^2+4x) +3(y^2-8y) + 62 \\ &=& 2[(x+2)^2-4] +3[(y-4)^2-16] + 62 \\ &=& 2(x+2)^2-8 +3(y-4)^2-48 + 62 \\ &=& 2(x+2)^2+3(y-4)^2-8 -48 + 62 \\ &=& 2(x {\color{red}+2})^2+3(y{\color{red}-4})^2 + 6\\ \hline \end{array}\)

We find global minimum:
\(\min\{ 2 x^2 + 3 y^2 + 8 x - 24 y + 62 \} = 6 \quad \text{at} \quad (x, y) = ( -2, 4)\)

1. What is the smallest whole number that has
a remainder of 1 when divided by 4,
a remainder of 1 when divided by 3,
and a remainder of 2 when divided by 5?

\(\begin{array}{|rcll|} \hline x &\equiv& 1 \pmod{4} \\ x &\equiv& 1 \pmod{3} \\ x &\equiv& 2 \pmod{5} \\ \hline \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline x &=& 1\cdot 3 \cdot 5 \cdot \frac{1}{3 \cdot 5}\pmod{4} + 1\cdot 4 \cdot 5 \cdot \frac{1}{4 \cdot 5}\pmod{3} + 2\cdot 4 \cdot 3 \cdot \frac{1}{4 \cdot 3}\pmod{5} + 4\cdot 3 \cdot 5 \cdot n \ |\ n\in Z \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline && \frac{1}{3 \cdot 5}\pmod{4} \\ &\equiv& (3 \cdot 5 )^{\varphi(4)-1} \pmod{4} \quad & | \quad \varphi(4) =4\cdot (1-\frac12) = 2 \\ &\equiv& 15^{1} \pmod{4} \\ &\equiv& 15 \pmod{4} \\ &\equiv& -1 \pmod{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{4 \cdot 5}\pmod{3} \\ &\equiv& (4 \cdot 5 )^{\varphi(3)-1} \pmod{3} \quad & | \quad \varphi(3) = 2 \\ &\equiv& 20^{1} \pmod{3} \\ &\equiv& 20 \pmod{3} \\ &\equiv& -1 \pmod{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{4 \cdot 3}\pmod{5} \\ &\equiv& (3 \cdot 5 )^{\varphi(4)-1} \pmod{4} \quad & | \quad \varphi(4) =4\cdot (1-\frac12) = 2 \\ &\equiv& 15^{3} \pmod{5} \quad & | \quad 12 \pmod{5} \equiv 2 \pmod{5} \\ &\equiv& 2^{3} \pmod{5} \\ &\equiv& 8 \pmod{5} \\ &\equiv& 3 \pmod{5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x &=& 15 \cdot (-1) + 20 \cdot (-1) + 24 \cdot 3 + 60 n \quad|\quad n\in Z \\ x &=& -15-20+72+60n \quad|\quad n\in Z \\ \mathbf{x} & \mathbf{=} & \mathbf{37+60n \quad|\quad n\in Z } \\ \hline \end{array}\)

\(\mathbf{x_{min} = 37 \quad | \quad n=0} \)

The smallest whole number is 37

12^15 =15,407,021,574,586,368

15,407,021,574,586,368 = 2^30 * 3^15 [45 factors - 2 distinct]

We have two geometric sequences of positive real numbers:

6,a,b and 1/b,a,54

Solve for a

I pick two whole numbers $x$ and $y$ between $1$ and $10$ inclusive (not necessarily distinct). My friend picks two numbers $x -4$ and $2y-1$. If the product of my friend's numbers is one greater than the product of my numbers, then what is the product of my numbers?

\(x = 9 \text{ and } y = 6 \\ xy = 54\)

The product of my numbers is 54.

There are 3 complex numbers $a+bi$, $c+di$, and $e+fi$. If $b=1$, $e=-a-c$,

and the sum of the numbers is $-i$, find $d+f$.

\(\begin{array}{|rcll|} \hline a+bi \\ c+di \\ e+fi \\ \hline a+c+e+(b+d+f)i &=& -i \quad & | \quad e=-a-c \\ a+c-a-c+(b+d+f)i &=& -i \\ (b+d+f)i &=& -i \quad & | \quad b = 1 \\ (1+d+f)i &=& -i \quad & | \quad :i \\ 1+d+f &=& -1 \\ \mathbf{d+f} &\mathbf{=}& \mathbf{-2} \\ \hline \end{array}\)

The quadratic $x^2-20x+36$ can be written in the form $(x+b)^2+c$, where $b$ and $c$ are constants.

What is $b+c$?

\(\begin{array}{|rclrcl|} \hline x^2+20x+36 &=& (x+b)^2+c \\ &=& x^2+\underbrace{2b}_{=20}\cdot x + \underbrace{(b^2+c)}_{=36} \\ \hline 2b &=& 20 & b^2+c &=& 36 \\ \mathbf{b} &\mathbf{=}& \mathbf{10} & 10^2+c &=& 36 \\ &&& c &=& 36 - 100 \\ &&& \mathbf{c} &\mathbf{=}& \mathbf{-64} \\\\ b+c &=& 10-64 \\ \mathbf{b+c} &\mathbf{=}& \mathbf{-54}\\ \hline \end{array}\)

If $x^2 + bx + b + 3 = 0$ has roots of the form $\frac{-b \pm \sqrt{5}}{2}$, where $b > 0 $,

then $b = m+\sqrt{n}$ for positive integers $m,n$. Find $m + n$.

\(\begin{array}{|rcll|} \hline \left( x - \frac{-b +\sqrt{5}}{2} \right)\left( x - \frac{-b - \sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( x + \frac{b}{2} -\frac{\sqrt{5}}{2} \right)\left( x + \frac{b}{2} +\frac{\sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( ( x + \frac{b}{2}) -\frac{\sqrt{5}}{2} \right)\left( (x + \frac{b}{2}) +\frac{\sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( x + \frac{b}{2} \right)^2- \left(\frac{\sqrt{5}}{2} \right)^2 &=& x^2+bx+b+3 \\ x^2+xb+\frac{b^2}{4}- \frac{5}{4} &=& x^2+bx+b+3 \\ \frac{b^2}{4}- \frac{5}{4} &=& b+3 \\ \frac{b^2}{4}-b - \frac{5}{4} -3 &=& 0 \quad & | \quad \cdot 4 \\ b^2-4b - 5 -12 &=& 0 \\ b^2-4b - 17 &=& 0 \\\\ b &=& \frac{4\pm \sqrt{16-4\cdot(-17)}}{2} \\ &=& \frac{4\pm \sqrt{16+68}}{2} \\ &=& \frac{4\pm \sqrt{84}}{2} \\ &=&2\pm \frac{\sqrt{84}}{2} \\ &=&2\pm \sqrt{\frac{84}{4}} \\ &=&2\pm \sqrt{21} \quad & | \quad b>0\ !\\\\ \mathbf{b} & \mathbf{=} & \mathbf{2+\sqrt{21} } \\ b &=& m+\sqrt{n} \quad & | \quad m = 2 \text{ and } n = 21 \\ \mathbf{m+n} &\mathbf{=}& \mathbf{2+21=23} \\ \hline \end{array}\)

We have two geometric sequences of positive real numbers:

6,a,b and 1/b,a,54

Solve for a

\(\text{Let $a_n = a_1q^{n-1}$} \\ \text{Let $a_{n-1} = a_1q^{n-2}$} \\ \text{Let $a_{n+1} = a_1q^{n}$}\)

\(\begin{array}{|rcll|} \hline a_{n-1}a_{n+1} &=& a_1q^{n-2}a_1q^{n} \\ &=& a_1^2q^{n-2}q^{n} \\ &=& a_1^2q^{n+n-2} \\ &=& a_1^2q^{2n-2} \\ &=& a_1^2q^{2(n-1)} \\ &=& a_1^2(q^{n-1})^2 \\\\ \sqrt{a_{n-1}a_{n+1}} &=& a_1 (q^{n-1}) \\ &=& a_n \\ \hline \end{array}\)

Formula:
\(\begin{array}{|rcll|} \hline a_n = \sqrt{a_{n-1}a_{n+1}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & a &=& \sqrt{6b} \\ (2) & a &=& \sqrt{\frac1b\cdot 54} \\\\ & a=\sqrt{6b} &=& \sqrt{\frac1b\cdot 54} \\ & 6b &=& \frac1b\cdot 54 \\ & b^2 &=& \frac{54}{6} \\ & b^2 &=& 9 \\ & \mathbf{ b } & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)

\(a=\ ?\)

\(\begin{array}{|rcll|} \hline a &=& \sqrt{6b} \quad & | \quad b= 3 \\ a &=& \sqrt{6\cdot 3} \\ a &=& \sqrt{2\cdot 3^2} \\ \mathbf{a} &\mathbf{=}& \mathbf{3\sqrt{2}} \\ \hline \end{array}\)

√9216 = 96

first we can make this into the square root of 289*25 which simplifies into 5 sqrt 289 which also simplifies into 5 * 17 (because 17^2=289) 5*17=85 which is our answer

Not to worry! I've found this:

Find the following limit:

lim_(x->2^-) (2 x)/(x^2 - 4)

If none of the terms of a product approach 0 as x->a, then by the product rule, the limit of the product is the product of the limits.

Applying the product rule, write lim_(x->2^-) (2 x)/(x^2 - 4) as 2 (lim_(x->2^-) x) (lim_(x->2^-) 1/(x^2 - 4)):

2 lim_(x->2^-) x lim_(x->2^-) 1/(x^2 - 4)

The limit of a continuous function at a point is just its value there.

lim_(x->2^-) x = 2:

2×2 lim_(x->2^-) 1/(x^2 - 4)

If lim_(x->2^-) y = 0, then the limit of 1/y is ± ∞.

Since lim_(x->2^-) (x^2 - 4) = 0 and x^2 - 4<0 for all x just to the left of x = 2, lim_(x->2^-) 1/(x^2 - 4) = -∞:

2×2 -∞

= -∞

In your equation you assumed that the length of the square prism is equal to s (the side of the base), isn't correct as if these lines were equal then the shape would be a cube rather than a square prism, nonetheless using my differing aprroach doesn't give an answer, so I assume the question wasn't written properly, as you need to assume the square prism is a cube to solve for l.

Are you sure it is $3.20 not $3.30? Because at $3.20 you can't have an EQUAL number of quarters and nickels.

1) Try 37.

To solve the equation for x....

\(|x|=x^2+x-3\\~\\ x=\pm(x^2+x-3)\\~\\ \begin{array}\ x=+(x^2+x-3)&\qquad\text{or}\qquad& x=-(x^2+x-3)\\~\\ x=x^2+x-3&&x=-x^2-x+3\\~\\ 0=x^2-3&&0=-x^2-2x+3\\~\\ 3=x^2&&x^2+2x-3=0\\~\\ \pm\sqrt3=x&&(x+3)(x-1)=0\\~\\ x=\sqrt3\quad\text{or}\quad x=-\sqrt3&\qquad\text{or}& x=-3\quad\text{or}\quad x=1 \end{array} \)

Now we need to test each possible solution to see if it makes the given equation true:

\(|\sqrt3|\,\stackrel?=\,\sqrt3^2+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3+\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,\sqrt3\)

Yes, so  \(x=\sqrt3\)  is a solution.

\(|-\sqrt3|\,\stackrel?=\,(-\sqrt3)^2-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,3-\sqrt3-3\\~\\ \sqrt3\,\stackrel?=\,-\sqrt3 \)

No, so  \(x=-\sqrt3\)  is not a solution.

\(|-3|\,\stackrel?=\,(-3)^2-3-3\\~\\ 3\,\stackrel?=\,9-3-3\\~\\ 3\,\stackrel?=\,3\)

Yes, so  \(x=-3\)  is a solution.

\(|1|\,\stackrel?=\,1^2+1-3\\~\\ 1\,\stackrel?=\,1+1-3\\~\\ 1\,\stackrel?=\,-1\)

No, so  \(x=1\)  is not a solution.

So the solutions are  \(x=\sqrt3\)  and  \(x=-3\)  .

AS = 5, 3, 1, -1, -3............etc

IS = 2^5, 2^3, 2^1, 2^-1, 2^-3

IS = 32, 8, 2, 1/2, 1/8........etc

8 / 32 = 1 / 4 - Common Ratio

Sum=F / [1 - R], where F=First term, R=Common ratio

Sum =32 / [1 - 1/4]

Sum =32 / [3/4]

Sum =32 x 4/3

Sum =42 2/3 - Sum of infinite series "B"

Thanks, I'n just glad I can help. 

Amazing! Beautiful, Gavin! 

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

The vector in the opposite direction of  \(\vec u\)  is  \(-\vec u\)  .

The unit vector in the same direction of  \(\vec u\)  is  \(\frac{\vec u}{|\vec u|}\)  which is  \(\frac{\vec u}{7}\)  .

So the unit vector in the opposite direction of  \(\vec u\)  is  \(-\frac{\vec u}{7}\)  .

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

Thank you very much for your help, you really helped me understand how to do these probelms. You're wonderful!