All Answers 358090 Answers

kay! I'll be there in a jiffy! haha.

Haha,Feel free to do so, lol

Aw. That's nice! I'll comee visit ;)

Haha

When grandma comes over= No School!!!

Thanks, Hun! & No school!! What! Lucky!

Okie Dokie! Hiya!

Thx Haley! I hope u hav a great Friday too! No school today!!!:D

Are you still looking for an answer? Please respond.

Im here!!! :)

1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?

\(\small{ \begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1} && 6 && 20 && 51 && 189 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 14 && 31 && 138 && 328 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 9} && 17 && 107 && 190 && 1352 && \cdots \\ \text{3. Difference } {\color{red}d_3 = 8} \\ \text{4. Difference } {\color{red}d_4 = 82} \\ \text{5. Difference } {\color{red}d_5 = -89} \\ \text{6. Difference } {\color{red}d_6 = 1175} \\ \text{7. Difference } {\color{red}d_7 = -4908} \\ \text{8. Difference } {\color{red}d_8 = 19363} \\ \end{array} }\)

\(\small{ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ &+& \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } + \binom{n-1}{5}\cdot {\color{red}d_5 }\\ &+& \binom{n-1}{6}\cdot {\color{red}d_6 } + \binom{n-1}{7}\cdot {\color{red}d_7 } + \binom{n-1}{8}\cdot {\color{red}d_8 }\\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 1 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 9 } \\ &+& \binom{n-1}{3}\cdot {\color{red} 8 } + \binom{n-1}{4}\cdot {\color{red} 82 } + \binom{n-1}{5}\cdot {\color{red} (-89) } \\ &+& \binom{n-1}{6}\cdot {\color{red} 1175 } + \binom{n-1}{7}\cdot {\color{red} (-4908) } + \binom{n-1}{8}\cdot {\color{red} 19363 } \\\\ a_{10} &=& \binom{9}{0}\cdot {\color{red} 1 } + \binom{9}{1}\cdot {\color{red} 5 } + \binom{9}{2}\cdot {\color{red} 9 } \\ &+& \binom{9}{3}\cdot {\color{red} 8 } + \binom{9}{4}\cdot {\color{red} 82 } + \binom{9}{5}\cdot {\color{red} (-89) } \\ &+& \binom{9}{6}\cdot {\color{red} 1175 } + \binom{9}{7}\cdot {\color{red} (-4908) } + \binom{9}{8}\cdot {\color{red} 19363 } \\ a_{10} &=& 1\cdot {\color{red} 1 } + 9\cdot {\color{red} 5 } + 36\cdot {\color{red} 9 } \\ &+& 84\cdot {\color{red} 8 } + 126\cdot {\color{red} 82 } + 126\cdot {\color{red} (-89) } \\ &+& 84\cdot {\color{red} 1175 } + 36\cdot {\color{red} (-4908) } + 9\cdot {\color{red} 19363 } \\ a_{10} &=& 1 + 45 + 324 +672 + 10332 -11214 +98700 - 176688+174267 \\ \mathbf{a_{10}} & \mathbf{=}& \mathbf{96439} \end{array} }\)

See you later then...:)

Heureka, could you please explain how you got \(2\leq n\leq14\) and \(a>0\) Where did those numbers come from?

(92 + 92 + 92 + x + x)/5 = 95

SOlve for x  276 +2x = 475

                             2x = 199

                                x = 99.5    

Lol InjustaGod, *rolls eyes* I got bored and left...:) No one was talking:)

Try to solve some of this questions

http://www.ime.eb.br/arquivos/Admissao/Vestibular_CFG/Provas_Anteriores/provas14_15/2014_Matematica_IME.pdf

Hi Juriemagic,

Admin knows that you and others have log in problems.  He does know how to fix it.

Have you tried using a different browser?  

Do you completely understand my answer?

You should thank guest answerer too.  He/she put in a lot of work on your behalf.   :/

Here are some more examples to practic on if you want

55555

- Bertie

Hi Melody!,

Thank you sooo much. I managed to do this sum but only to a point...

Yes, the logging on is still a problem, aw well, I'll speak with I.T. here and see if they can help...otherwise?..I trust you are all okay?

AAwwww..so easy!!!!!..thanx a million!

Well you obviously don't have all night, ur not on

You can fit an order 8 polynomial to these numbers.  The coefficients look awful but the next term given by the polynomial is 96439.

I cant find the angle between the vertices c (1,1,1) a (1,-4,2) b (-5,2,7) Can you help me?

I set:

\(\small{ \begin{array}{lcr} a_x &=& 1\\ a_y&=&-4\\ a_z&=&2 \end{array} \qquad \begin{array}{lcr} b_x &=& -5\\ b_y&=&2\\ b_z&=&7 \end{array} \qquad \begin{array}{lcr} c_x &=& 1\\ c_y &=& 1\\ c_z &=& 1\\ \end{array} }\)

Differences of the vectors:
\(\small{ \begin{array}{lcr} \vec{b}-\vec{a} &=& \bigl(\begin{smallmatrix} b_x-a_x\\b_y-a_y\\b_z-a_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr)\\ \vec{a}-\vec{b} &=& -\bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{c}-\vec{b} &=& \bigl(\begin{smallmatrix} c_x-b_x\\c_y-b_y\\c_z-b_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr)\\ \vec{b}-\vec{c} &=& -\bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{a}-\vec{c} &=& \bigl(\begin{smallmatrix} a_x-c_x\\a_y-c_y\\a_z-c_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)\\ \vec{c}-\vec{a} &=& -\bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \\ \end{array} }\)

In the plane of the triangle (ABC) the angles are:

\(\small{ \begin{array}{rcl} \tan{(A)} &=& \frac{ \vert (\vec{c}-\vec{a}) \times (\vec{b}-\vec{a}) \vert} {(\vec{c}-\vec{a})\cdot (\vec{b}-\vec{a}) } = \frac{ \left| \bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {25 } \qquad (I.)\\ A &=& \arctan{(1.74218253923)}\\ A &=& 60.1444921468^\circ \\ \tan{(B)} &=& \frac{ \vert (\vec{a}-\vec{b}) \times (\vec{c}-\vec{b}) \vert} {(\vec{a}-\vec{b})\cdot (\vec{c}-\vec{b}) } = \frac{ \left| \bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {72 } \qquad (I.)\\ B &=& \arctan{(0.60492449279)}\\ B &=& 31.1707710617^\circ \\ \tan{(C)} &=& \frac{ \vert (\vec{b}-\vec{c}) \times (\vec{a}-\vec{c}) \vert} {(\vec{b}-\vec{c})\cdot (\vec{a}-\vec{c}) } = \frac{ \left| \bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)} = \frac{ 43.5545634808 } {1 } \qquad (I.)\\ C &=& \arctan{(43.5545634808)}\\ C &=& 88.6847367915^\circ \\ \end{array} }\)

\(A+B+C = 60.1444921468^\circ + 31.1707710617^\circ + 88.6847367915^\circ = 180^\circ \)

give your whats app no.

If we have

\((x+3)^2=0\)

then the only value for x is -3

Now \((x+3)^2\rightarrow x^2+6x+9\)

so m is 6 (and n is 9)

.

Like so:

\(\frac{(4x^2y)(-2xy^3)}{(-2xy)^4}\rightarrow \frac{-8x^3y^4}{16x^4y^4}\rightarrow-\frac{1}{2x}\)