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Let the maximum number of people who rode the bike all 3 days =S
S + (15 - S) + (12 - S) + (9 - S) = 22
-2S + 36 = 22
-2S = 22 - 36
-2S = -14
S = -14 / -2

S = 7 - Maximum number of people.

The solution in the fourth quadrant is 1 - sqrt(3)/3*i.

Find a formula for the exponential function passing through the points  (-1,2/3) and (1,6) Help much apprieciated can't figure this one out.

The answer is 7/2.

1) theta = 0, pi/2, 3*pi/2, 2*pi

2) theta = pi/4, pi, 3*pi/4

The answer is 1/4018020.

For the first problem, the answer is (8,6) and (4,-2).

For the second problem, the answer is (-5,5).

The answer is 17.

The answer is 2010.

What is the remainder when   x^3 + x^6 + x^9 + x^27   is divided by   x^2 - 1  ?     

Well answered by Heureka and Tiggsy.   Thanks.

Let F_1 = ( -3, 1 - sqrt(5)/4) and F_ 2= ( -3, 1 + sqrt(5)/4). Then the set of points P$ such that |PF_1 - PF_2| = 1 form a hyperbola. The equation of this hyperbola can be written as ((y - k)^2)/(a^2) - ((x - h)^2)/(b^2) = 1,\]where a, b > 0. Find h + k + a + b.

Thanks Tiggsy,

That makes good sense :)

Let

 \(\displaystyle f(x)=x^{3}+x^{6}+x^{9}+x^{27}\quad \text{and}\quad \frac{f(x)}{x^{2}-1}=Q(x)+\frac{R(x)}{x^{2}-1}\\ \text{where R(x) is linear, i.e.}\; R(x)=ax+ b \quad \text{for some}\; a\; \text{and}\; b.\\ \text{Then}\\ f(x)=Q(x)(x^{2}-1)+R(x)\\ \text{so}\\ f(-1)=-2=R(-1)=-a+b\dots\dots(1)\\ f(1)= 4=R(1)=a+b\dots\dots\dots(2)\)

and from (1) and (2), a = 3 and b = 1.

Two of the cars are chosen at random.

What is the probability that the two cars differ in all three categories?

I assume,

\(\begin{array}{|rcll|} \hline && { 24\times 1 \times 2 \times 3}\above 1pt \dfrac{24!}{(24-2)!} \\\\ &=& \dfrac{144}{552} \\\\ &=& 0.26086956522\ \approx 26\% \\ \hline \end{array}\)

Thanks very much Tiggsy.  I have not studied it yet but I will.  

See diagram from previous post.

Triangles BEO and CGO are congruent.

Let BE = x , the angle at E is a right angle so  x = (R + h)cos(theta).

EA = 5 - x, and if AD is to be a tangent to the semi circle, then the line from A to the point of tangency will have length 5 - x also.

Similarly on the RHS, the line from D to the point of tangency will have length 4 - x.

So, if AD is to be a tangent to the semi circle, it will have a length 9 - 2x = 9 - 2(R + h)cos(theta).

Now move into co-ordinate geometry mode.

Take BC to be the horizontal axis, with O as the origin.

Point A will have an x co-ordinate   -(R + h) + 5cos(theta) and a y co-ordinate 5sin(theta).

Point D will have an x co-ordinate    (R + h) - 4cos(theta) and a y co-ordinate 4sin(theta).

So,

\(\displaystyle AD^{2}= \{9-2(R+h)\cos\theta\}^{2}=\{(R+h)-4\cos\theta-\{-(R+h)+5\cos\theta\}\}^{2}+(4\sin\theta-5\sin\theta)^{2},\\ 81-36(R+h)\cos\theta+4(R+h)^{2}\cos^{2}\theta=4(R+h)^{2}-36(R+h)\cos\theta+81\cos^{2}\theta+\sin^{2}\theta,\\ 81+4(R+h)^{2}\cos^{2}\theta-4(R+h)^{2}-81\cos^{2}\theta-\sin^{2}\theta = 0, \\ 4(R+h)^{2}(\cos^{2}\theta-1)+81\sin^{2}\theta -\sin^{2}\theta=0,\\ 4(R+h)^{2}(-\sin^{2}\theta)+80\sin^{2}\theta=0,\\ (R+h)^{2}=20,\\ R+h =\sqrt{20}=2\sqrt{5} . \)

\(\displaystyle \text{BC}=2(R+h)=4\sqrt{5}.\)

Thanks Heureka,

What is the remainder when x^3 + x^6 + x^9 + x^27 is divided by x^2 - 1?

Heureka has done this the standard way but I just wanted to see if I could figure out how to do it without the 

algebraic division.  I think my method is valid but I am not 100% sure on the divide by 2 bit.  I think it is valid though.

\(x^2-1=(x-1)(x+1)\\ let\;\;f(x)=x^3+x^6+x^9+x^{27}\\ \text{When f(x) is divided by x+1 the remainder will be }f(-1)\\ f(-1)=-1+1-1-1=-2\\ \frac{x^3+x^6+x^9+x^{27}}{(x+1)(x-1)}\\ =\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{(1+1+1+1+2)/2}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\\qquad \qquad \text{I divided by 2 because I factored out (x+1)}\\ \\ \qquad \qquad if \;\;Q(x)=x+1\;\;then\;\;Q(1)=2\\~\\ =g(x)+\frac{3}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{3(x+1)}{(x-1)(x+1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{3x+1}{(x-1)(x+1)}\\ \text{so the remainder will be }3x+1\)

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coding

x^2-1=(x-1)(x+1)\\
let\;\;f(x)=x^3+x^6+x^9+x^{27}\\
\text{When f(x) is divided by x+1 the remainder will be }f(-1)\\
f(-1)=-1+1-1-1=-2\\
\frac{x^3+x^6+x^9+x^{27}}{(x+1)(x-1)}\\
=\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{(1+1+1+1+2)/2}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\\qquad \qquad \text{I divided by 2 because I factored out (x+1)}\\
\\ \qquad \qquad if \;\;Q(x)=x+1\;\;then\;\;Q(1)=2\\~\\
=g(x)+\frac{3}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{3(x+1)}{(x-1)(x+1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{3x+1}{(x-1)(x+1)}\\
\text{so the remainder will be }3x+1

Express \(\cos(5x)\) as a polynomial in \(\cos(x)\).

\(\begin{array}{|rcll|} \hline \mathbf{\Big(\cos(x) + i\sin(x) \Big)^5} &=& \mathbf{\cos(5x) +i\sin(5x)} \\\\ \Big(\cos(x) + i\sin(x) \Big)^5 &=& \binom50\cos^5(x) \\ &+& \binom51\cos^4(x)(i)\sin(x) \\ &+& \binom52\cos^3(x)(i^2)\sin^2(x) \quad & | \quad i^2=-1 \\ &+& \binom53\cos^2(x)(i^3)\sin^3(x) \quad & | \quad i^3=-i \\ &+& \binom54\cos(x)(i^4)\sin^4(x) \quad & | \quad i^4=1 \\ &+& \binom55(i^5)\sin^5(x) \quad & | \quad i^5=i \\\\ &=& \binom50\cos^5(x) \\ &+& \binom51\cos^4(x)(i)\sin(x) \\ &-& \binom52\cos^3(x)\sin^2(x) \\ &-& \binom53\cos^2(x)(i)\sin^3(x) \\ &+& \binom54\cos(x)\sin^4(x) \\ &+& \binom55(i)\sin^5(x) \\ \hline \end{array} \)

Compare the real parts of each side:

\(\begin{array}{|rcll|} \hline \cos(5x) &=& \binom50\cos^5(x) \\ &-& \binom52\cos^3(x)\sin^2(x) \\ &+& \binom54\cos(x)\sin^4(x) \\\\ \cos(5x) &=& \binom50\cos^5(x) \quad & | \quad \binom50 = 1 \\ &-& \binom52\cos^3(x)\sin^2(x) \quad & | \quad \binom52 = 10,\ \sin^2(x)=1-\cos^2(x) \\ &+& \binom54\cos(x)\sin^2(x)\sin^2(x) \quad & | \quad \binom54 = 5,\ \sin^2(x)=1-\cos^2(x) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x)\Big(1-\cos^2(x)\Big) \\ &+& 5\cos(x)\Big(1-\cos^2(x)\Big)\Big(1-\cos^2(x)\Big) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x)\Big(1-\cos^2(x)\Big)^2 \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x)\Big(1-2\cos^2(x)+\cos^4(x)\Big) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x) -10\cos^3(x)+5\cos^5(x) \\\\ \mathbf{\cos(5x)} &=& \mathbf{ 16\cos^5(x) - 20\cos^3(x) + 5\cos(x) } \\ \hline \end{array}\)

Solve the equation \(9^{5-x}= \dfrac{1}{\sqrt{15^{4x}}}\).

Express your answer in terms of \(\ln3\) and \(\ln5\) .

\(\begin{array}{|rcll|} \hline 9^{5-x} &=& \dfrac{1}{\sqrt{15^{4x}}} \\ 9^{5-x} &=& \dfrac{1}{15^{\frac{4x}{2}}} \\ 9^{5-x} &=& \dfrac{1}{15^{2x}} \\ 9^59^{-x} &=& \dfrac{1}{15^{2x}} \\ 9^53^{-2x} &=& \dfrac{1}{15^{2x}} \\ 9^5 &=& \dfrac{3^{2x}}{15^{2x}} \\ 9^5 &=& \left(\dfrac{3}{15}\right)^{2x} \\ 9^5 &=& \left(\dfrac{1}{5}\right)^{2x} \\ 9^5 &=& \dfrac{1}{5^{2x}} \\ 9^55^{2x} &=& 1 \\ 3^{2\cdot 5}5^{2x} &=& 1 \\ 3^{10}5^{2x} &=& 1 \\ 5^{2x} &=& 3^{-10} \quad | \quad \ln \text{ both sides } \\ 2x\ln(5) &=& -10\ln(3) \quad | \quad : 2 \\ x\ln(5) &=& -5\ln(3) \quad | \quad : 2 \\\\ \mathbf{x} &=& \mathbf{ -\dfrac{5\ln(3)}{\ln(5)} } \\ \hline \end{array}\)

The first two positive integers \(n\) for which \(1 + 2 + \ldots + n\) is a perfect square are \(1\) and \(8\). 

What are the next two?

Definition:

\(\text{Triangular numbers: $a(n) = \dbinom{n+1}{2} =\dfrac{ n(n+1)}{2} = 1 + 2 +3+4+ \ldots + n$. }\\\\ 1, 3, 6, 10, 15, 21, 28, 36,\ldots \)

\(\text{$a(m)$-th triangular number is a square: $\\\mathbf{a(m+1) = 6*a(m)-a(m-1)+2}$, with $a(1) = 1$, $a(2) = 8$. } \\\\ 1, 8, 49, 288, 1681, 9800, 57121, 332928, \ldots \)

Example:

\(\begin{array}{|c|r|r|r|} \hline & a(m+1) & & s_n \\ m+1 & = 6*a(m)-a(m-1)+2 & n & = \dfrac{ n(n+1)}{2} \\ \hline 3 & 6*a(2)-a(1)+2 \\ & = 6*8-1+2 \\ & = 49 & 49 & 35^2 \\\\ \hline 4 & 6*a(3)-a(2)+2 \\ & 6*49-8 +2 \\ & 288 & 288 & 204^2 \\\\ \hline 5 & 6*a(4)-a(3)+2 \\ & 6*288-49 +2 \\ & 1681 & 1681 & 1189^2 \\\\ \hline \ldots \\ \hline \end{array} \)

The next two positive integers n are 49 and 288.

I got 1/5 

There are 125 smaller cubes, each with 6 sides, so there are 750 total faces. Each face of the original cube has a 5x5 square of blue faces, so there are 25 blue faces for each side of the original cube, for a total of 150 blue faces. 150/750=1/5