heureka

what is 1/2+1/4+1/8

\(\begin{array}{|rcll|} \hline && \frac12 + \frac14 + \frac18 \\ &=& \frac12 + \frac{1}{2^2} + \frac{1}{2^3} \\ &=& \frac12 \cdot ( 1 + \frac{1}{2} + \frac{1}{2^2} ) \\ &=& \frac12 \cdot [ 1 + \frac{1}{2} \cdot ( 1+ \frac{1}{2} ) ] \\ &=& \frac12 \cdot ( 1 + \frac{1}{2} \cdot \frac{3}{2} ) \\ &=& \frac12 \cdot ( 1 + \frac{3}{4} ) \\ &=& \frac12 \cdot \frac{7}{4} \\ &=& \frac{7}{8} \\ \hline \end{array}\)

-138 mod 116

\(\begin{array}{|rcll|} \hline -138 \equiv x \pmod {116} \\ -138+116 \equiv x \pmod {116} \quad & | \quad 116 \pmod {116} = 0 \\ -22+116 \equiv x \pmod {116} \quad & | \quad 116 \pmod {116} = 0 \\ 94 \equiv x \pmod {116} \\\\ -138 \pmod {116} = 94 \pmod {116} = 94 \\ \hline \end{array} \)

what is the surface area of a cylinder with the radius of 5 and height of 11

Let A = surface area
Let r = radius
Let h = height

\(\begin{array}{|rcll|} \hline A &=& (2\pi r)\cdot h + 2\cdot (\pi r^2) \\ A &=& (2\pi r)\cdot (h + r) \quad & | \quad r = 5 \qquad h = 11 \\ A &=& (2\pi 5)\cdot (11 + 5) \\ A &=& 10\cdot \pi\cdot 16 \\ A &=& 160\cdot \pi \\ A &=& 502.654824574 \\ \hline \end{array}\)

3600 + 9*2150 + 17*1300 + 16*775 + 18*460 + 68*275 + 65*165 + 19*100 = 97055

Please help me find the next three numbers in this geometric sequence

625,225,75...  

and please explain how you got that answer so i can understand.

A geometric  sequence has the same factor  between each number. Not here !

\(\frac{625}{225} = 2.77777777778 \\ \frac{225}{75} = 3 \\ 2.77777777778 \ne 3 \)

xpand by using Binomial theorem (5-x)^6

\(\begin{array}{|rcll|} \hline && (5-x)^6 \\ &=& [(-1)\cdot (x-5)]^6 \\ &=& (-1)^6 \cdot (x-5)^6 \quad & | \quad (-1)^6 = 1 \\ &=& (x-5)^6 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && (x-5)^6 \\ &=& \binom60 x^6\cdot 5^0 - \binom61 x^5\cdot 5^1 + \binom62 x^4\cdot 5^2 - \binom63 x^3\cdot 5^3 \\ && + \binom64 x^2\cdot 5^4 - \binom65 x^1\cdot 5^5 + \binom66 x^0\cdot 5^6 \\ &=& \binom60 x^6 - 5^1\cdot \binom61 x^5 + 5^2\cdot \binom62 x^4 - 5^3\cdot \binom63 x^3 \\ && + 5^4\cdot \binom64 x^2 - 5^5\cdot \binom65 x^1 + 5^6 \cdot \binom66\\ && \begin{array}{|rcll|} \hline \binom60 = \binom66 = 1 \\ \binom61 = \binom65 = 6 \\ \binom62 = \binom64 = 15 \\ \binom63 = 20 \\ \hline \end{array} \\ &=& 1\cdot x^6 - 5^1\cdot 6 \cdot x^5 + 5^2\cdot 15 \cdot x^4 - 5^3\cdot 20 \cdot x^3 \\ && + 5^4\cdot 15\cdot x^2 - 5^5\cdot 6 \cdot x^1 + 5^6 \cdot 1\\ &=& x^6 - 30\ x^5 +375\ x^4-2500\ x^3 + 9375\ x^2-18750\ x+15625 \\ \hline \end{array} \)

Solve for X

\(\begin{bmatrix} 4 && -4 && 7 \\ 8 && 2 && 1 \end{bmatrix} +X= \begin{bmatrix} -3 && 4 && 6\\ -17 && 9 && 18 \end{bmatrix}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} 4 && -4 && 7 \\ 8 && 2 && 1 \end{bmatrix} +X &=& \begin{bmatrix} -3 && 4 && 6\\ -17 && 9 && 18 \end{bmatrix} \\ X &=& \begin{bmatrix} -3 && 4 && 6\\ -17 && 9 && 18 \end{bmatrix} -\begin{bmatrix} 4 && -4 && 7 \\ 8 && 2 && 1 \end{bmatrix} \\ X &=& \begin{bmatrix} -3-4 && 4-(-4) && 6-7 \\ -17-8 && 9-2 && 18-1 \end{bmatrix} \\ \mathbf{ X } & \mathbf{=} & \mathbf{ \begin{bmatrix} -7 && 8 && -1 \\ -25 && 7 && 17 \end{bmatrix} }\\ \hline \end{array} \)

I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.

\(\int_{-\pi/3}^{0}(\sqrt{\cos x})(\sin^{3}x) dx\)

Until \(-1 \int_{\cos -\pi /3}^{\cos 0} (u)^{1/2}(1-u^{2})du\) it is correct.

distribute and find the endpoints:

\(\begin{array}{|rcll|} \hline && -\int \limits_{\frac12}^{1} u^{\frac12}(1-u^{2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12}\cdot u^{2})\ du \quad & | \quad u^{\frac12}*u^2 \text { is not }u !!!\\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+\frac42})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\color{red}{\frac52}})\ du \\ &=& - \left[\frac23 u^{\frac32}- \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& \left[-\frac23 u^{\frac32}+ \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& -\frac23\cdot (1) + \frac27 \cdot(1)-[-\frac23\cdot (\frac12)^{\frac32}+\frac27\cdot (\frac12)^{\frac72} ] \\ &=& -\frac23 + \frac27 +\frac{1}{3\sqrt{2}}-\frac{1}{28\sqrt{2}} \quad & | \quad \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ &=& -\frac23 + \frac27 +\frac{\sqrt{2}}{6}-\frac{\sqrt{2}}{56} \\ &=& \frac{-14+6}{21} + \frac{56-6}{6\cdot 56}\cdot \sqrt{2} \\ &=& \frac{-8}{21} + \frac{50}{336}\cdot \sqrt{2} \\ &=& -\frac{8}{21} + \frac{25}{168}\cdot \sqrt{2} \\ \hline \end{array}\)

how do i find the constant term in the expansion of (2x^2-1/x)^6

\(\begin{array}{|rcll|} \hline && \left( 2x^2-\frac{1}{x} \right)^6 \\\\ &=& \left( \frac{2x^3-1}{x} \right)^6 \\\\ &=& \frac{(2x^3-1)^6}{x^6} \\\\ &=& \frac{1}{x^6} \cdot (2x^3-1)^6 \\ \hline \end{array} \)

The constant  term is at the 5th term in the binomial expansion:

\(\begin{array}{|rcll|} \hline && \frac{1}{x^6} \cdot \binom64 \cdot (2x^3)^2 \quad & | \quad \binom64 = \binom{6}{6-4} = \binom62\\ &=& \frac{1}{x^6} \cdot \binom62 \cdot (2x^3)^2 \\ &=& \frac{1}{x^6} \cdot \binom62 \cdot 4x^6 \\ &=& 4 \cdot \binom62 \\ &=& 4 \cdot \frac62\cdot \frac51 \\ &=& 4 \cdot 3\cdot 5 \\ &=& 60 \\ \hline \end{array} \)

Deterimine whether the line 5x+12y=169 is a tangent to the circl  x2 + y2=169. if so, find the point where the tangent line touches the circle.

Discussion:
If the line and the circle have no intersections, then the line is not a tangent.
If the line and the circle have one intersection, then the line is a tangent.
If the line and the circle have two intersections, then the line is not a tangent.

We compute the intersections:

\(\begin{array}{|lrcll|} \hline (1) & 5x+12y &=& 169 \quad & | \quad 169 = 13^2 \\ & 5x+12y &=& 13^2 \\ & 12y &=& 13^2 -5x \\ & y &=& \frac{13^2 -5x}{12} \\\\ (2) & x^2 +y^2 &=& 169 \quad & | \quad 169 = 13^2 \\ & x^2 +y^2 &=& 13^2 \quad & | \quad y = \frac{13^2 -5x}{12} \\ & x^2 + (\frac{13^2 -5x}{12})^2 &=& 13^2 \\ & x^2 + \frac{(13^2 -5x)^2}{12^2} &=& 13^2 \quad & | \quad \cdot 12^2\\ & 12^2x^2 + (13^2 -5x)^2 &=& 12^2\cdot 13^2 \\ & 12^2x^2 + 13^2\cdot 13^2 -2\cdot 13^2\cdot 5x+5^2x^2 &=& 12^2\cdot 13^2 \\ & 12^2x^2 + 5^2x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \\ & (12^2+5^2)x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \quad & | \quad 12^2+5^2=13^2 \\ & 13^2x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \quad & | \quad : 13^2 \\ & x^2 + 13^2 -10x &=& 12^2 \quad & | \quad - 12^2 \\ & x^2 -10x + 13^2 - 12^2 &=& 0 \quad & | \quad 13^2 - 12^2=5^2 \\\\ & x^2 -10x + 5^2 &=& 0 \\ & x^2 -10x + 25 &=& 0 \\\\ & x &=& \frac{10\pm \sqrt{100-4\cdot 25} }{2} \\ & x &=& \frac{10\pm \sqrt{100-100} }{2} \\ & x &=& \frac{10\pm 0 }{2} \\ & x &=& \frac{10}{2} \\ & \mathbf{ x } & \mathbf{=} & \mathbf{5} \\\\ & y &=& \frac{13^2 -5x}{12} \\ & y &=& \frac{13^2 -5^2}{12} \quad & | \quad 13^2 -5^2=12^2 \\ & y &=& \frac{12^2}{12} \\ & \mathbf{ y } & \mathbf{=} & \mathbf{12} \\ \hline \end{array} \)

We have only one intersection at x = 5 and y = 12,

so the line touches the circle.