heureka

The function f(x) = ax^r satisfies

f(2) = 1 and

f(32) = 4.
Find r.

\(\begin{array}{|lrclclr|} \hline & \mathbf{f(x)} & \mathbf{=} & \mathbf{ax^r} \\\\ x=2 : & f(2) &=& a\cdot 2^r &=& 1 & (1) \\ x=32: & f(32) &=& a\cdot 32^r &=& 4 & (2) \\\\ \hline (2) \div (1) : & \frac{a\cdot 32^r} { a\cdot 2^r } &=& \frac{4}{1} \\ & \frac{ 32^r} { 2^r } &=& 4 \\ & \left( \frac{ 32 } { 2 } \right)^r &=& 4 \\ & 16^r &=& 4 \quad &&& | \quad 16 = 4^2 \qquad 4 = 4^1 \\ & 4^{2r} &=& 4^1 \\ & 2r &=& 1 \quad &&& | \quad : 2 \\ & \mathbf{r} & \mathbf{=} & \mathbf{ \frac12 } \\ \hline \end{array}\)

Hey, I just need help with this problem. I'm not sure about the working out, can someone please help me out?

Differentiate\( \frac{2x^5-3x^4+6x^3-2x^2 }{3x^2}\)

The answer is \(2x^2-2x+2\)

Thank you 

\(\begin{array}{|rcll|} \hline y &=& \frac{2x^5-3x^4+6x^3-2x^2 }{3x^2} \\ &=& \frac{x^2 \cdot \left( 2x^3-3x^2+6x-2\right) }{3x^2} \\ &=& \frac{ 2x^3-3x^2+6x-2 }{3} \\ &=& \frac23 x^3-x^2+2x-\frac23 \\\\ y' & = & \frac23\cdot 3x^2 -2x + 2 \\ \mathbf{y'} & \mathbf{=} & \mathbf{2x^2 -2x + 2} \\ \hline \end{array}\)

define polar axis symmetry and its tests for the graph r=f(theta)

Let \(m\) and \(n\) denote the greatest and least positive three-digit multiples of 7, respectively.

What is the value of \(m + n\) ?

\(\begin{array}{|rcll|} \hline m &=& 7\cdot \left \lfloor{ \frac{999}{7} }\right \rfloor{} \\ &=& 7\cdot 142 \\ &=& 994 \\\\ n &=& 7\cdot \left \lceil{ \frac{100}{7} }\right \rceil{} \\ &=& 7\cdot 15 \\ &=& 105 \\\\ m+n &=& 994 + 105 \\ &=& 1099 \\ \hline \end{array}\)

5^2*5^4*5^6*.......5^(2n) = (0.008)^(-30) find the value of n

\(\begin{array}{|rclrcl|} \hline 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 0.008^{(-30)} \\ && 0.008 &=& \frac{8}{1000} \\ && &=& \frac{2^3}{10^3} \\ && &=& \left( \frac{2}{10} \right)^3 \\ && &=& \left( \frac{1}{5} \right)^3 \\ && &=& \frac{1}{5^3} \\ && &=& 5^{-3} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& (5^{-3})^{-30} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{(-3)\cdot (-30) } \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{90} \\ 5^{2+4+6 \cdot \ldots \cdot 2n } &=& 5^{90} \\\\ 2+4+6 \cdot \ldots \cdot 2n &=& 90 \\ 2\cdot(1+2+3 + \ldots + n ) &=& 90 \\ && 1+2+3 + \ldots + n &=& \frac{(1+n)\cdot n}{2} \\ 2\cdot \Big(\frac{(1+n)\cdot n}{2} \Big) &=& 90 \\ (1+n)\cdot n &=& 90 \\ n^2+n -90 &=& 0 \\ (n-9)\cdot(n+10) &=& 0 \\ \hline \end{array}\)

\(n = 9 \text{ or } n = -10\)

Because n > 0:

n = 9

\(5^2\cdot 5^4 \cdot 5^6\cdot 5^8\cdot 5^{10}\cdot 5^{12} \cdot 5^{14}\cdot 5^{16}\cdot 5^{18} = 0.008^{(-30)} \)

Let n be a positive integer and a be an integer such that a is its own inverse modulo n.
What is the remainder when a^2 is divided by n?

A modular inverse of an integer a (modulo n) is the integer \(a^{(-1)}\) such that
\(aa^{(-1)}=1 \pmod n\)

So let \(a^{(-1)} = a\), we have:

\(\begin{array}{|rcll|} \hline a\cdot a &=& 1 \pmod n \\ a^2 &=& 1 \pmod n \\ \hline \end{array}\)

The remainder when a² is divided by n ist 1

GIVEN:

a and b are real numbers

 \(a+b = 10 \\ a^2+b^2 = 44\)

Find \(a^3 + b^3\)

\(\begin{array}{|rcll|} \hline (a+b)^2 &=& a^2 + 2ab + b^2 \\ (a+b)^2 &=& (a^2 + b^2) +2ab \\ (10)^2 &=& (44) + 2ab \\ 100 &=& 44 + 2ab \quad & | \quad : 2 \\ 50 &=& 22 + ab \\ ab &=& 50-22 \\ \mathbf{ab} & \mathbf{=} & \mathbf{28} \\\\ (a^2+b^2)(a+b) &=& a^3 +a^2b + b^2a + b^3 \\ (a^2+b^2)(a+b) &=& a^3 +b^3 + ab(a+b) \\ 44\cdot 10 &=& a^3 +b^3 + 28\cdot 10 \\ 440 &=& a^3 +b^3 + 280 \\ a^3 +b^3 &=& 440 - 280 \\ \mathbf{ a^3 +b^3} & \mathbf{=} & \mathbf{160} \\ \hline \end{array}\)

A Chinese emperor orders a regiment of soldiers in his palace to divide into groups of 4.
They do so successfully.
He then orders them to divide into groups of 3, upon which 2 of them are left without a group.
He then orders them to divide into groups of 11, upon which 5 are left without a group.
If the emperor estimates there are about two hundred soldiers in the regiment,
what is the most likely number of soldiers in the regiment?

\(\begin{array}{|rcll|} \hline n &\equiv& {\color{red}0} \pmod {{\color{green}4}} \\ n &\equiv& {\color{red}2} \pmod {{\color{green}3}} \\ n &\equiv& {\color{red}5} \pmod {{\color{green}11}} \\\\ m &=& {\color{green}4}\cdot {\color{green}3} \cdot {\color{green}11} \\ &=& 132 \\ \hline \end{array} \)

\(\text{Because } 3\cdot 11 \text{ and } 4 \text{ are relatively prim } ( gcd(33,4) = 1! ) \\ \text{and because } 4\cdot 11 \text{ and } 3 \text{ are relatively prim } ( gcd(44,3) = 1! ) \\ \text{and because } 4\cdot 3 \text{ and } 11 \text{ are relatively prim } ( gcd(12,11) = 1! ) \\ \text{we can continue}\)

\(\small{ \begin{array}{|rcll|} \hline n &=& {\color{red}0} \cdot {\color{green}3\cdot 11} \cdot \Big( \frac{1}{\color{green}3\cdot 11}\pmod {{\color{green}4}} \Big) \\ & +& {\color{red}2} \cdot {\color{green}4\cdot 11} \cdot \Big( \frac{1}{\color{green}4\cdot 11}\pmod {{\color{green}3}} \Big) \\ & +& {\color{red}5} \cdot {\color{green}4\cdot 3} \cdot \Big( \frac{1}{\color{green}4\cdot 3}\pmod {{\color{green}11}} \Big) \\ & +& {\color{green}4}\cdot {\color{green}3} \cdot {\color{green}11} \cdot k \quad & | \quad k\in Z\\\\ n &=& {\color{red}2} \cdot {\color{green}44} \cdot \Big( \frac{1}{\color{green}44}\pmod {{\color{green}3}} \Big) +{\color{red}5} \cdot {\color{green}12} \cdot \Big( \frac{1}{\color{green}12}\pmod {{\color{green}11}} \Big) +132 \cdot k \quad & | \quad k\in Z\\ \hline \end{array} } \)

\(\begin{array}{rcll} && \frac{1}{\color{green}44}\pmod {{\color{green}3}} \quad &| \quad \text { modular inverse } 44\cdot \frac{1}{44} \equiv 1 \pmod {3} \\ &=& {\color{green}44}^{\varphi({\color{green}3})-1} \pmod {{\color{green}3}} \quad & | \quad \text{ Euler's totient function } \varphi(n) \quad \varphi(p) = p-1 \\ &=& 44^{2-1} \pmod {3} \\ &=& 44 \pmod {3} \\ &=& 2 \\\\ && \frac{1}{\color{green}12}\pmod {{\color{green}11}} \quad &| \quad \text { modular inverse } 12\cdot \frac{1}{12} \equiv 1 \pmod {11} \\ &=& {\color{green}12}^{\varphi({\color{green}11})-1} \pmod {{\color{green}11}} \quad & | \quad \text{ Euler's totient function } \varphi(n)\quad \varphi(p) = p-1 \\ &=& 12^{10-1} \pmod {11} \\ &=& 12^{9} \pmod {11} \quad & | \quad 12 \equiv 1 \pmod {11 } \\ &=& 1^{9} \pmod {11} \\ &=& 1 \\ \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline n &=& {\color{red}2} \cdot {\color{green}44} \cdot \Big( 2 \Big) +{\color{red}5} \cdot {\color{green}12} \cdot \Big( 1 \Big) +132 \cdot k \\ &=& 176 +60 +132 \cdot k \\ &=& 236 +132 \cdot k \quad & | \quad 236 \equiv 104 \pmod {132} \\ \mathbf{n} & \mathbf{=} & \mathbf{104 +132 \cdot k } \quad & | \quad \mathbf{ k\in Z } \\ \hline \end{array} } \)

\(\begin{array}{|lrcll|} \hline k=0: & n & = & 104 \\ k=1: & n & = & 104+132 \\ & &\mathbf{=}& \mathbf{236} \\ k=2: & n & = & 104+132\cdot 2 \\ & & = & 368 \\ \ldots \\ \hline \end{array} \)

The most likely number of soldiers in the regiment is 236

Find the area of a triangle with side lengths frac43, 2, and frac83

Let \(a = \frac43\)

Let \(b = 2\)

Let \(c = \frac83\)

\(\begin{array}{rcll} \mathbf{Formula:} \\ \hline \text{Area of a triangle with Heron} \\ \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ s &=& \frac{a+b+c}{2} \\ \end{array} \)

\(\begin{array}{lcll} \mathbf{s = \ ? } \\ \begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \\ &=& \frac{\frac43+2+\frac83}{2} \\ &=& \frac23+1+\frac43 \\ &=& \frac63+1 \\ &=& 2+1 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array} \\ \end{array} \)

\(\begin{array}{lcll} \mathbf{A = \ ? } \\ \begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ &=&\sqrt{3(3-\frac43)(3-2)(3-\frac83)} \\ &=&\sqrt{3(\frac{9}{3}-\frac43)(1)(\frac{9}{3}-\frac83)} \\ &=&\sqrt{3(\frac{9-4}{3})(\frac{9-8}{3})} \\ &=&\sqrt{3(\frac{5}{3})(\frac{1}{3})} \\ &=&\sqrt{5(\frac{1}{3})} \\ &\mathbf{=}&\mathbf{\sqrt{\frac53}} \\ \hline \end{array} \\ \end{array}\)

The area is \( \mathbf{\sqrt{\frac53} = 1.29}\)

If a,b,c are integers from the set of positive integers less than 7 such that

\(\begin{align*} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align*}\)

then what is the remainder when a+b+c is divided by 7 ?

\(\begin{array}{rcll} \mathbf{Formula}: \\\\ 6b & \equiv & 3+b \pmod 7 \quad & | \quad -b \\ 6b-b & \equiv & 3+b-b \pmod 7 \\ 5b & \equiv & 3 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{5b} &\mathbf{\equiv}& \mathbf{3 \pmod 7} \\ \hline \end{array} & (1) \\\\ \begin{array}{|rcll|} \hline \mathbf{5c} &\mathbf{\equiv}& \mathbf{2 \pmod 7} \\ \hline \end{array} & (2) \\\\ \begin{array}{|rcll|} \hline \mathbf{abc} &\mathbf{\equiv}& \mathbf{1 \pmod 7} \\ \hline \end{array} & (3) \\\\ \begin{array}{|rcll|} \hline \mathbf{a+b+c} &\mathbf{\equiv}& \mathbf{x \pmod 7} \\ \hline \end{array} & (4) \\\\ (1)+(2): \qquad 5b+5c & \equiv & 3+2 \pmod 7\\ 5(b+c) & \equiv & 5 \pmod 7 \quad & | \quad :5 \\ \frac{5(b+c)}{5} & \equiv & \frac{5}{5} \pmod 7 \\ b+c & \equiv & 1 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{b+c} &\mathbf{\equiv}& \mathbf{1 \pmod 7} \\ \hline \end{array} & (5) \\\\ (5) \text{ in }(4): \qquad a+b+c & \equiv & x \pmod 7 \\ a+ (b+c) & \equiv & x \pmod 7 \\ a+ (1) & \equiv & x \pmod 7 \\ a+ 1 & \equiv & x \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{a+ 1 } &\mathbf{\equiv}& \mathbf{x \pmod 7} \\ \hline \end{array} & (6) \\\\ (1) \times (2): \qquad 5b\times 5c & \equiv & 3\times 2 \pmod 7 \\ 25bc & \equiv & 6 \pmod 7 \quad & | \quad 6 \equiv -1 \pmod 7 \\ 25bc & \equiv & -1 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{25bc} &\mathbf{\equiv}& \mathbf{-1 \pmod 7} \\ \hline \end{array} & (7) \\\\ \mathbf{a=\ ? } \\ (3) \div (7): \qquad \frac{abc}{25bc} & \equiv & \frac{1}{-1} \pmod 7 \\ \frac{a}{25} & \equiv & -1 \pmod 7 \quad & | \quad \times 25 \\ \frac{25a}{25} & \equiv & (-1)\cdot 25 \pmod 7 \\ a & \equiv & -25 \pmod 7 \\ a & \equiv & -25 + 4\times 7 \pmod 7 \\ a & \equiv & -25 + 28 \pmod 7 \\ a & \equiv & 3 \pmod 7 \\ a-3 &=& n\cdot 7 \\ a &=& 3+ n\cdot 7 \qquad n \in \mathbb{Z} \\ \begin{array}{|rcll|} \hline \mathbf{a} &\mathbf{=}& \mathbf{ 3+ n\cdot 7 \qquad n \in \mathbb{Z} } \\ \hline \end{array} & (8) \\\\ \mathbf{x=\ ? } \\ (6): \qquad a+1 & \equiv & x \pmod 7 \quad & | \quad a = 3+ n\cdot 7 \\ 3+ n\cdot 7 + 1 & \equiv & x \pmod 7 \\ 4+ n\cdot 7 & \equiv & x \pmod 7 \quad & | \quad n\cdot 7 \pmod 7 = 0 \\ 4 & \equiv & x \pmod 7 \\ 4 \pmod 7 &=& x \\ 4 &=& x \\ \begin{array}{|rcll|} \hline \mathbf{x} &\mathbf{=}& \mathbf{ 4 } \\ \hline \end{array} \\ \end{array}\)

The remainder when \(a+b+c\) is divided by \(7\) is \(\mathbf{4}\)