heureka

Thank you Melody.

what is the next 2 terms -9 -8 -4 5 21 46

$\begin{array}{|rcll|} \hline a_n &=& -9 + \underbrace{\frac{(n-1)\cdot n \cdot (n-\frac12 )}{3}}_{\text{sum of the square numbers}} \quad n = 1,2,3, \cdots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_7 &=& -9 + \frac{(7-1)\cdot 7 \cdot (7-\frac12 )}{3} \\ a_7 &=& -9 + \frac{6\cdot 7 \cdot (\frac{14}{2}-\frac12 )}{3} \\ a_7 &=& -9 + \frac{6\cdot 7 \cdot (\frac{13}{2})}{3} \\ a_7 &=& -9 + 7 \cdot 13 \\ \mathbf{a_7 }& \mathbf{=} & \mathbf{82} \\\\ a_8 &=& -9 + \frac{(8-1)\cdot 8 \cdot (8-\frac12 )}{3} \\ a_8 &=& -9 + 7\cdot 4 \cdot 5\\ \mathbf{a_8} & \mathbf{=} & \mathbf{131}\\ \hline \end{array}$

Can someone help me use the definition of a derivative using the f(x) listed below.

$f(x)=\frac{1}{\sqrt{x^2+5}}$

$\begin{array}{|rcll|} \hline f'(x) &=& \lim \limits_{h\to 0} \frac{ f(x+h) - f(x) } {h} \\\\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \left( \frac{1}{ \sqrt{(x+h)^2+5} } - \frac{1}{ \sqrt{x^2+5}} \right) \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \quad | \quad (a-b)(a+b) = a^2-b^2\\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \left( \frac{1}{ (x+h)^2+5 } - \frac{1}{ x^2+5 } \right) } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \frac{(x^2+5)-[(x+h)^2+5] }{ [(x+h)^2+5]\cdot(x^2+5) } } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \frac{(x^2+5)-(x^2+2xh+h^2+5) }{ [(x+h)^2+5]\cdot(x^2+5) } } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \frac{(x^2+5)-x^2-2xh-h^2-5) }{ [(x+h)^2+5]\cdot(x^2+5) } } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \frac{x^2+5-x^2-2xh-h^2-5 }{ [(x+h)^2+5]\cdot(x^2+5) } } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ \frac{ -2xh-h^2 }{ [(x+h)^2+5]\cdot(x^2+5) } } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ -2xh-h^2 } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) \cdot [(x+h)^2+5]\cdot(x^2+5) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ h\cdot(-2x-h) } {h \cdot \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) \cdot [(x+h)^2+5]\cdot(x^2+5) } \\ f'(x) &=& \lim \limits_{h\to 0} \frac{ -2x-h } { \left( \frac{1}{ \sqrt{(x+h)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) \cdot [(x+h)^2+5]\cdot(x^2+5) } \\ f'(x) &=& \frac{ -2x-0 } { \left( \frac{1}{ \sqrt{(x+0)^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) \cdot [(x+0)^2+5]\cdot(x^2+5) } \\ f'(x) &=& \frac{ -2x } { \left( \frac{1}{ \sqrt{x^2+5} } + \frac{1}{ \sqrt{x^2+5}} \right) \cdot (x^2+5)\cdot(x^2+5) } \\ f'(x) &=& \frac{ -2x } { 2\cdot \frac{1}{ \sqrt{x^2+5} } \cdot (x^2+5)^2 } \\ f'(x) &=& \frac{ - x } { \frac{1}{ \sqrt{x^2+5} } \cdot (x^2+5)^2 } \\ f'(x) &=& \frac{ -x } { \frac{(x^2+5)^2}{ \sqrt{x^2+5} } } \\ f'(x) &=& \frac{ -x } { \frac{ \sqrt{(x^2+5)^4}} { \sqrt{x^2+5} } } \\ f'(x) &=& \frac{ -x } { \sqrt{ \frac{ (x^2+5)^4 } { x^2+5} } } \\ f'(x) &=& \frac{ -x } { \sqrt{ (x^2+5)^3 } } \\ \hline \end{array}$

five consecutive terms of an arithmetic sequence have a sum of 40. The product of the first, middle and the last term is 224. Find the terms of sequence PLEASE HELP !

I.

$\begin{array}{|rcll|} \hline a_i + a_{i+1}+ a_{i+2}+ a_{i+3}+ a_{i+4} &=& 40 \\ a_i + (a_i+d) + (a_i+2d)+ (a_i+3d)+ (a_i+4d) &=& 40 \\ 5a_i + 10d &=& 40 \qquad & | \qquad :5\\ a_i + 2d &=& 8 \\ \mathbf{a_i} & \mathbf{=} & \mathbf{ 8 - 2d }\\ \hline \end{array}$

II.

$\begin{array}{|rcll|} \hline a_i \cdot a_{i+2} \cdot a_{i+4} &=& 224 \\ a_i \cdot (a_i+2d) \cdot (a_i+4d) &=& 224 \qquad & | \qquad a_i = 8 - 2d \\ (8 - 2d) \cdot (8 - 2d+2d) \cdot (8 - 2d+4d) &=& 224 \\ (8 - 2d) \cdot 8 \cdot (8 + 2d) &=& 224 \qquad & | \qquad :8 \\ (8 - 2d) \cdot (8 + 2d) &=& \frac{224}{8} \\ (8 - 2d) \cdot (8 + 2d) &=& 28 \qquad & | \qquad (u-v)\cdot (u+v)=u^2-v^2 \\ 8^2-(2d)^2 &=& 28 \\ 64-4d^2 &=& 28 \\ 4d^2 &=& 64-28 \\ 4d^2 &=& 36 \qquad & | \qquad :4 \\ d^2 &=& 9 \\ d &=& \sqrt{9} \\ \mathbf{d} & \mathbf{=} & \mathbf{ 3 }\\ \hline \end{array}$

III.

$\begin{array}{|rcll|} \hline a_i &=& 8-2d \qquad & | \qquad d = 3 \\ a_i &=& 8-2\cdot 3 \\ a_i &=& 8-6 \\ \mathbf{a_i} & \mathbf{=} & \mathbf{ 2 }\\ \hline \end{array}$

five consecutive terms of an arithmetic sequence: $\dots, 2, 5, 8, 11, 14, \dots$

$\begin{array}{|lrcll|} \hline \text{sum: } & 2+5+8+11+14 = 40 \\ \text{product: } & 2\cdot 8 \cdot 14 = 224 \\ \hline \end{array}$

0.6464... (repeating) as a Fraction?

$\begin{array}{|rcll|} \hline 1\cdot a &=& 0.6464... \text{ (repeating)} \\ 100\cdot a &=& 64.6464... \text{ (repeating)}) \\\\ 100\cdot a - 1\cdot a &=& 64.6464... \text{ (repeating)} - 0.6464... \text{ (repeating)}\\ 99\cdot a &=& 64 \\ a &=& \frac{64}{99} \\ \mathbf{ 0.6464... \text{ (repeating)} } &\mathbf{=}& \mathbf{ \frac{64}{99} }\\ \hline \end{array}$

can someone please help me find the derivative of this:
$f(x)=\sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}}$

$\begin{array}{|rcllll|} \hline f(x) &=& \sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}} \\ && & \sqrt[3]{x} = x^{\frac13} \\ && & \frac{2}{{x}^{2016}} = 2 \cdot x^{-2016} \\ && & \frac{1}{\sqrt{x^3}} = \frac{ 1 } { x^{\frac32} } = x^{-\frac32} \\ f(x) &=& x^{\frac13} + 2 \cdot x^{-2016} + x^{-\frac32} \\ && & \text{Formula: } \\ && & y = x^n \\ && & y' = n\cdot x^{n-1} \\\\ f'(x) &=& \frac13 \cdot x^{\frac13-1} + 2 \cdot (-2016) \cdot x^{-2016-1} + (-\frac32) x^{-\frac32-1} \\ f'(x) &=& \frac13 \cdot x^{-\frac23 } -4032 \cdot x^{-2017} -\frac32\cdot x^{-\frac52} \\ && & x^{-\frac23 } = \frac{ 1 } { x^{\frac23} } = \frac{1}{\sqrt[3]{x^2}} \\ && & x^{-2017} = \frac{1}{{x}^{2017}} \\ && & x^{-\frac52 } = \frac{ 1 } { x^{\frac52} } = \frac{1}{\sqrt{x^5}}\\ f'(x) &=& \frac13 \cdot \frac{1}{\sqrt[3]{x^2}} -4032 \cdot \frac{1}{{x}^{2017}} -\frac32\cdot \frac{1}{\sqrt{x^5}} \\ \mathbf{f'(x)} & \mathbf{=} & \mathbf{\frac{1}{3 \sqrt[3]{x^2}} - \frac{4032}{{x}^{2017}} - \frac{3}{2 \sqrt{x^5}} }\\ \hline \end{array}$

Your answer is the same and also correct.

(a-b+c-d)(a+b-c+d)

$\begin{array}{|rcll|} \hline && (a-b+c-d)(a+b-c+d) \\ &=& [a-(b-c+d)][a+(b-c+d)] \qquad | \qquad (u-v)(u+v) = u^2-v^2 \\ &=& a^2-(b-c+d)^2 \\ &=& a^2-(b-c+d)(b-c+d) \\ &=& a^2-[(b-c)+d][(b-c)+d] \qquad | \qquad (u+v)(u+v) = u^2+2uv+v^2\\ &=& a^2-[ (b-c)^2+2(b-c)d+d^2 ] \\ &=& a^2-(b-c)^2-2(b-c)d-d^2 \\ &=& a^2-(b-c)^2-2bd+2cd-d^2 \qquad | \qquad (u-v)(u-v) = u^2-2uv+v^2\\ &=& a^2-(b^2-2bc+c^2)-2bd+2cd-d^2 \\ &=& a^2-b^2+2bc-c^2-2bd+2cd-d^2 \\ &=& a^2-b^2-c^2-d^2+2bc-2bd+2cd \\ \hline \end{array}$

Please Help

Example: Found the team with index 2:

 $\begin{array}{|lcll|} \hline 1. \text{Search: team index } \le 16 \qquad | \qquad \text{yes} \\ 2. \text{Search: team index } \le 8 \qquad | \qquad \text{yes} \\ 3. \text{Search: team index } \le 4 \qquad | \qquad \text{yes} \\ 4. \text{Search: team index } \le 2 \qquad | \qquad \text{yes} \\ 5. \text{Search: team index } \le 1 \qquad | \qquad \text{no} \\ \mathbf{6.} \text{Search: team index } = 2 \qquad | \qquad \text{yes and found} \\ \hline \end{array}$

In the worst case 6 items. At most, 6.

Write a rule for a quadratic function with a graph that has x-intercepts (2,0) and (-6,0) and a maximum point of (-2,4)

quadratic function: $y =ax^2+bx+c$

1. Point (x=2, y=0):

$\quad 0 = a\cdot (2^2)+b\cdot 2 + c \\ \Rightarrow 4a+2b+c=0$

Maximum Point ($x_{max} = -2, \ y_{max} = 4$):

$\begin{array}{|rcll|} \hline x_{max} &=& -\frac{b}{2a} \\ -2 &=& -\frac{b}{2a} \\ 2 &=& \frac{b}{2a} \\ 4a &=& b \\ \mathbf{ a } & \mathbf{=}& \mathbf{\frac{b}{4} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 4a+2b+c&=&0 \quad | \quad a=\frac{b}{4} \\ 4\cdot \frac{b}{4} +2b+c&=&0 \\ b+2b+c&=&0 \\ 3b+c&=&0 \\ 3b&=&-c \\ \mathbf{ b } & \mathbf{=}& \mathbf{-\frac{c}{3}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a &=& \frac{b}{4} \quad | \quad b = -\frac{c}{3} \\ a &=& \frac{ -\frac{c}{3}}{4} \\ a &=& -\frac{c}{12} \\\\ y_{max} &=& a\cdot ( x_{max} ) ^2 + b\cdot x_{max} + c \quad | \quad a = -\frac{c}{12} \quad b = -\frac{c}{3} \\ y_{max} &=& -\frac{c}{12}\cdot ( x_{max} ) ^2 -\frac{c}{3}\cdot x_{max} + c \quad | \quad x_{max} = -2\\ y_{max} &=& -\frac{c}{12}\cdot ( -2 ) ^2 -\frac{c}{3}\cdot (-2) + c \quad | \quad y_{max} = 4\\ 4 &=& -\frac{1}{3}\cdot c +\frac{2}{3}\cdot c + c \\ 4 &=& \frac{4}{3}\cdot c \\ \mathbf{ c } & \mathbf{=}& \mathbf{3} \\\\ b&=&-\frac{c}{3}\\ \mathbf{ b } & \mathbf{=}& \mathbf{-1} \\\\ a&=&-\frac{b}{4}\\ \mathbf{ a } & \mathbf{=}& \mathbf{-\frac{1}{4}} \\ \hline \end{array}$

quadratic function:

$\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \\ \mathbf{ y }&\mathbf{=}& \mathbf{-\frac{1}{4}x^2-x+3} \\ \hline \end{array}$