heureka

x^a=y^b=k and x^c=y^d=t, then which of the following is true.

a) ac=bd

b) ad=bc

c) a/d=c/b

d) a^c=b^d

e) a+c=b+d

\(x^a=y^b \\ x^c=y^d\)

\(\begin{array}{|rcll|} \hline x^a &=& y^b \quad & | \quad \text{exponentiate both sides with } \frac{c}{a} \\ \displaystyle x^{a\cdot\frac{c}{a}} &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ x^{c} &=& \displaystyle y^{b\cdot\frac{c}{a}}\quad & | \quad x^{c} = y^d \\ y^d &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ \Rightarrow d &=&\displaystyle b\cdot\frac{c}{a} \\ \mathbf{ad} & \mathbf{=} & \mathbf{bc} \\ \hline \end{array} \)

The values of the four variables a, b, c, and d are 9, 11, 13, and 15, though not necessarily in that order.

What is the number of possible values of the expression ab+bc+cd+da?

Let a = 9
Let b = 11
Let c = 13
Let d = 15

\(\small{ \begin{array}{|r|c|r|r|r|r|rcl|} \hline & \text{All permutations} & & & & & \\ & \text{of a,b,c,d} & A & B & C & D & AB+BC+CD+DA &=& (B+D)(A+C) \\ \hline 1. & abcd & 9 & 11 & 13 & 15 & (11+15)(9+13) = 26*22 &=& 572 \\ 2. & abdc & 9 & 11 & 15 & 13 & (11+13)(9+15) = 24*24 &=& \qquad 576 \\ 3. & acbd & 9 & 13 & 11 & 15 & (13+15)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 4. & acdb & 9 & 13 & 15 & 11 & (13+11)(9+15) = 24*24 &=& \qquad 576 \\ 5. & adcb & 9 & 15 & 13 & 11 & (15+11)(9+13) = 26*22 &=& 572 \\ 6. & adbc & 9 & 15 & 11 & 13 & (15+13)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 7. & bacd & 11 & 9 & 13 & 15 & (9+15)(11+13) = 24*24 &=& \qquad 576 \\ 8. & badc & 11 & 9 & 15 & 13 & (9+13)(11+15) = 22*26 &=& 572 \\ 9. & bcad & 11 & 13 & 9 & 15 & (13+15)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 10. & bcda & 11 & 13 & 15 & 9 & (13+9)(11+15) = 22*26 &=& 572 \\ 11. & bdca & 11 & 15 & 13 & 9 & (15+9)(11+13) = 24*24 &=& \qquad 576 \\ 12. & bdac & 11 & 15 & 9 & 13 & (15+13)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 13. & cbad & 13 & 11 & 9 & 15 & (11+15)(13+9) = 26*22 &=& 572 \\ 14. & cbda & 13 & 11 & 15 & 9 & (11+9)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 15. & cabd & 13 & 9 & 11 & 15 & (9+15)(13+11) = 24*24 &=& \qquad 576 \\ 16. & cadb & 13 & 9 & 15 & 11 & (9+11)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 17. & cdab & 13 & 15 & 9 & 11 & (15+11)(13+9) = 26*22 &=& 572 \\ 18. & cdba & 13 & 15 & 11 & 9 & (15+9)(13+11) = 24*24 &=& \qquad 576 \\ 19. & dbca & 15 & 11 & 13 & 9 & (11+9)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 20. & dbac & 15 & 11 & 9 & 13 & (11+13)(15+9) = 24*24 &=& \qquad 576 \\ 21. & dcba & 15 & 13 & 11 & 9 & (13+9)(15+11) = 22*26 &=& 572 \\ 22. & dcab & 15 & 13 & 9 & 11 & (13+11)(15+9) = 24*24 &=& \qquad 576 \\ 23. & dacb & 15 & 9 & 13 & 11 & (9+11)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 24. & dabc & 15 & 9 & 11 & 13 & (9+13)(15+11) = 22*26 &=& 572 \\ \hline \end{array} }\)

The number of possible values of the expression ab+bc+cd+da=(b+d)(a+c) is 3

The values are 560, 572, and 576

integrals

5) \(\displaystyle \int^{\pi/2}_{0}\sin^{13}x\;dx\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\sin^{12}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\Big(\sin^2(x)\Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big(1-\cos^2(x) \Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big( \binom{6}{0}-\binom{6}{1}\cos^2(x) \\ && +\binom{6}{2}\cos^4(x)-\binom{6}{3}\cos^6(x)+\binom{6}{4}\cos^8(x) \\ && -\binom{6}{5}\cos^{10}(x)+\binom{6}{6}\cos^{12}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ \end{array}\)

Integration by parts
\(\begin{array}{rclrcl} u&=&\cos^n(x) & v &=& -\cos(x) \\ u'&=&n\cos^{n-1}(x)\sin(x) & v' &=& \sin(x) \\ \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\cos^n(x)}_{u}\underbrace{\sin(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\cos^n(x)}_{u}(\underbrace{-\cos(x)}_{v}) \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{-n \cos^{n-1}(x)\sin(x)}_{u'}(\underbrace{-\cos(x)}_{v})\;dx \\ &=& \Big[ -\cos^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\cos^{n}(x)\sin(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx &=& -\Big[ \cos^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx} & \mathbf{=} & \mathbf{\frac{-\Big[\cos^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array} }\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 6\cdot \frac{1}{3}+15\cdot \frac{1}{5} - 20\cdot \frac{1}{7}+15\cdot \frac{1}{9}- 6\cdot \frac{1}{11}+1\cdot \frac{1}{13} \\\\ &=& \mathbf{\dfrac{1024}{3003} } \\ \end{array}\)

integrals

4) \(\displaystyle \int^{\pi/2}_{0}\cos^{15}{x}\;dx\)

without substitution:

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\cos^{14}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\Big(\cos^2(x)\Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big(1-\sin^2(x) \Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big( \binom{7}{0}-\binom{7}{1}\sin^2(x) \\ && +\binom{7}{2}\sin^4(x)-\binom{7}{3}\sin^6(x)+\binom{7}{4}\sin^8(x) \\ && -\binom{7}{5}\sin^{10}(x)+\binom{7}{6}\sin^{12}(x)-\binom{7}{7}\sin^{14}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ \end{array}\)

Integration by parts

\(\begin{array}{rclrcl} u&=&\sin^n(x) & v &=& \sin(x) \\ u'&=&n\sin^{n-1}(x)\cos(x) & v' &=& \cos(x) \\ \end{array}\)

\(\begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\sin^n(x)}_{u}\underbrace{\cos(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\sin^n(x)}_{u}\underbrace{\sin(x)}_{v} \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{n \sin^{n-1}(x)\cos(x)}_{u'}\underbrace{\sin(x)}_{v}\;dx \\ &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\sin^{n}(x)\cos(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx} & \mathbf{=} & \mathbf{\frac{\Big[\sin^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array}\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 7\cdot \frac{1}{3}+21\cdot \frac{1}{5} - 35\cdot \frac{1}{7}+35\cdot \frac{1}{9}- 21\cdot \frac{1}{11}+7\cdot \frac{1}{13} - 1\cdot \frac{1}{15} \\\\ &=& \mathbf{\dfrac{2048}{6435} } \\ \end{array}\)

integrals

2) \(\displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}\left(1-\frac{\sqrt{x}}{\sqrt[3]{x}} \right) }\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x} }\;dx \\ \end{array} \)

For the integrand \(\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x}}\), substitute \(u = \sqrt[6]{x}\) or \(u^5 = x^{\frac{5}{6}}\) and

\(\begin{array}{rcll} du &=& \frac16 x^{\frac16-1} \;dx \\ &=& \frac16 x^{-\frac{5}{6}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{ x^{\frac{5}{6}}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{u^5} \;dx \\ \text{so } dx &=& 6u^5\;du \\ \end{array} \)

This gives a new lower bound \(u = \sqrt[6]{0} = 0\) and upper bound \(u = \sqrt[6]{1} = 1\)

\(\begin{array}{rcll} &=& \displaystyle \int \limits_{0}^{1}u\sqrt{1-u}\cdot 6u^5\;du \\ &=& 6\displaystyle \int \limits_{0}^{1}u^6\sqrt{1-u} \;du \\ \end{array} \)

For the integrand \(u^6\sqrt{1-u}\), substitute \(v = \sqrt{1-u}\)  or \(u = 1-v^2\) and

\(\begin{array}{rcll} dv &=& \dfrac{-1}{2\sqrt{1-u}} \;du \\ &=& \dfrac{-1}{2v} \;du \\ \text{so } du &=& -2v \;dv \\ \end{array}\)

This gives a new lower bound \(v = \sqrt{1} = 1\) and upper bound \(v = \sqrt{0} = 0\)

\(\begin{array}{rcll} &=& 6\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v\cdot (-2)v\;dv \\ &=& -12\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v^2\;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} (1-6v^2+15v^4-20v^6+15v^8-6v^{10}+v^{12} )v^2 \;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} ( v^2-6v^4+15v^6-20v^8+15v^{10}-6v^{12}+v^{14}) \;dv \\ &=& -12 \cdot \Big[ \frac{v^3}{3}-6\frac{v^5}{5}+15\frac{v^7}{7}-20\frac{v^9}{9}+15\frac{v^{11}}{11}-6\frac{v^{13}}{13}+\frac{v^{15}}{15} \Big]_{1}^{0} \\ &=& 12 \cdot\Big( \frac{1}{3}-\frac{6}{5}+\frac{15}{7}-\frac{20}{9}+\frac{15}{11}-\frac{6}{13}+\frac{1}{15} \Big) \\ &=& 12 \cdot\Big( \frac{675675-6\cdot405405+15\cdot289575-20\cdot225225+15\cdot184275-6\cdot 155925+135135}{2027025} \Big) \\ &=& \dfrac{552960}{2027025} \\ \\ &=& \dfrac{135\cdot 4096}{135\cdot 15015} \\ \\ &=& \dfrac{ 4096}{ 15015} \\ \end{array} \)

integrals

1)  \(\displaystyle \int \limits_{0}^{1}\sqrt{1-\sqrt{x}}\;dx\)

 For the integrand  \(\sqrt{1-\sqrt{x}} \) , substitute \(u = \sqrt{1-\sqrt{x}} \) or \(\sqrt{x}=1-u^2\) and

 \(\begin{array}{rcll} du &=& \frac12(1-\sqrt{x})^{\frac12-1} (-\frac12)x^{\frac12-1} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{u(1-u^2)} \;dx \\\\ \text{so } dx &=& -4u(1-u^2)\;du \end{array}\)

This gives a new lower bound \(u = \sqrt{1} = 1 \) and upper bound \(u = \sqrt{0} = 0\)
\(\begin{array}{rcll} &=& -4 \displaystyle \int \limits_{1}^{0}u\cdot u(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2-u^4\;du \\ &=& -4\cdot \Big( \displaystyle \int \limits_{1}^{0}u^2 \;du-\displaystyle \int \limits_{1}^{0} u^4\;du \Big) \\ &=& -4\cdot \Big( \Big[ \frac{u^3}{3} \Big]_{1}^{0} -\Big[ \frac{u^5}{5} \Big]_{1}^{0} \Big) \\ &=& -4\cdot \Big( -\frac{1}{3} - \Big( -\frac{1}{5} \Big) \Big) \\ &=& -4\cdot \Big( \frac{1}{5} -\frac{1}{3} \Big) \\ &=& -4\cdot \left( \dfrac{-2}{15} \right) \\ &=& \dfrac{8}{15} \\ \end{array} \)

what is the interior and exterior angles of a 21-agon

exterior angle \(= \dfrac{360^{\circ}}{21} = 17.1428571429^{\circ}\)

interior angle \( = 180^{\circ} -\) exterior angle \( = 180^{\circ} -17.1428571429^{\circ} = 162.857142857^{\circ}\)

​ Help with Log Question!

Solve for x.

\(\begin{array}{|rcll|} \hline \log_2(x-2)+\log_2(x+6) &=& 7 \quad & | \quad \log(a*b) = \log(a)+ \log(b) \\ \log_2(~(x-2)*(x+6)~) &=& 7 \quad & | \quad 2^{()} \\ (x-2)*(x+6) &=& 2^7 \\ (x-2)*(x+6) &=& 128 \\ (x-2)*(x+6) -128 &=& 0 \\ x^2+6x-2x-12-128 &=& 0 \\ x^2+4x -140 &=& 0 \\\\ x &=& \dfrac{-4\pm \sqrt{16-4*(-140)} }{2} \\ x &=& \dfrac{-4\pm \sqrt{16+4*140} }{2} \\ x &=& \dfrac{-4\pm \sqrt{576} }{2} \\ x &=& \dfrac{-4\pm 24 }{2} \\ x &=& -2\pm 12 \\\\ x_1 &=& -2+ 12 \\ \mathbf{x_1} & \mathbf{=}& \mathbf{10} \\\\ x_2 &=& -2- 12 \\ \mathbf{x_2} & \mathbf{=}& \mathbf{-14} \\ && x_2 \text{ is no solution,}\\ && \text{because } x_2 < 0 ! \\ \hline \end{array}\)

A positive integer has exactly eight positive divisors. If two of the divisors are 21 and 35, what is the number? NOTE: For some integer n, a divisor of n is a non-zero integer that divides exactly into n leaving a remainder of 0. For example: 3 is a divisor of 18 since, 18 ÷ 3 = 6 remainder 0. 4 is not a divisor of 18, since 18 ÷ 4 = 4 remainder 2.

lcm(21,35) = 105

John weighed the pumpkins two at a time. After he is done weighing the pumpkins he observed the following weights:

110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 pounds. 
How much did each pumpkin weigh? 
Explain Clearly how you got the answer?