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Thank you Heureka! 

EDIT: why did you say help at the top of your answer? It just caught my eye.

-\(tommarvoloriddle\)

Hallo Heureka, your name is hard to spell, thank you a ton, how do you make a diagram in Microsoft word?🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️ and why does this feel like a mirror on the wall thing? It kinda feels like it rhymes.

-\(tommarvoloriddle\)

B)
What is \(\sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)}\)?

\(\begin{array}{|rcll|} \hline && \mathbf{ \sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)} } \\\\ &=& \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{ k} - \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=4}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=1}^{\infty} \dfrac{1}{ k+3} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{3} \dfrac{1}{ k} \\\\ &=& \dfrac13 \left( \dfrac{1}{ 1}+\dfrac{1}{ 2}+\dfrac{1}{ 3}\right) \\\\ &=& \dfrac13 \left( \dfrac{6+3+2}{ 6} \right) \\\\ &=& \dfrac13 \left( \dfrac{11}{ 6} \right) \\\\ &=& \mathbf{ \dfrac{11}{ 18} } \\ \hline \end{array}\)

a 0.112 kg apple on a tree has a PE of 4.00 J. How high is it above the ground?

Hello Guest!

\(F=\frac{G\cdot m\cdot M}{r^2}\\ r=\sqrt{\frac{G\cdot m\cdot M}{F}}\)

F = PE = 4J

m = 0.112 kg

M = \(5.97\cdot 10^{24}\)kg

G = \(\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\)

\(r=\sqrt{\frac{\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\cdot 0.112kg\cdot 5.97\cdot 10^{24}kg}{4\frac{kg\cdot m}{s^2}}}\)

\(r=\sqrt{\frac{m^3\cdot kg\cdot kg\cdot s^2}{kg\cdot s^2\cdot kg\cdot m}\times \frac{6.672\cdot 10^{-11}\cdot 0.112\cdot 5.97\cdot 10^{24}}{4}}\)

\(r=\sqrt{9152\cdot 10^{13}\ m^2}\\ r=3339598.05965\ m\)

\(high=3339598.05965\ m-6378 m (Earth's\ radius, geocentric \ mean)\\ \color{blue} high=3333220\ m\)

The tree stands on the south pole and is about 3333221m high.

Sorry, I read 4 N instead of 4 J.

\(PE= m\times g\times h\\ h=\frac{PE}{m\times g}\\ h=\frac{4J\times s^2}{0.112kg\times 9,81m}\times\frac{kg\times m^2}{J\times s^2}\\ \color{blue}h=3.641m\)

  !

A)
There exist real numbers \(A\) and \(B\) so that \(\dfrac{1}{k(k + 3)} = \dfrac{A}{k} + \dfrac{B}{k + 3}\).
for all real numbers \(k\) other than \(0\) and \(-3\). Enter the ordered pair \((A,B)\).

\(\begin{array}{|lrcll|} \hline &\mathbf{\dfrac{1}{k(k + 3)}} &=& \mathbf{\dfrac{A}{k} + \dfrac{B}{k + 3}} \quad | \quad \cdot k(k + 3) \\\\ &1 &=& \dfrac{Ak(k + 3)}{k} + \dfrac{Bk(k + 3)}{k + 3} \\\\ &1 &=& A (k + 3) + Bk \\ \hline k=-3 &1 &=& A (-3 + 3) + B(-3) \\ &1 &=& 0 -3B \\ &1 &=& -3B \\ & \mathbf{B} &=& \mathbf{-\dfrac{1}{3}} \\ \hline k=0 &1 &=& A (0 + 3) + B(0) \\ &1 &=& 3A+ 0 \\ &1 &=&3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array} \)

\(\mathbf{(A,B) = \left(\dfrac{1}{3},\ -\dfrac{1}{3} \right) }\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{k(k + 3)} &=& \dfrac{\frac{1}{3}}{k} + \dfrac{-\frac{1}{3}}{k + 3} \\ &=& \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \\ \hline \end{array} \)

Hallo tommarvoloriddle,

thank you,

i did make the diagram with Microsoft Word !

Help.

\(\begin{array}{|lrcll|} \hline (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline (1)-(2): & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- (a^2-2ab+b^2) \\ & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- a^2+2ab-b^2 \\ & (a+b)^2-(a-b)^2 &=& 4ab \\ & (a+b)^2 &=& 4ab+(a-b)^2 \\ & \mathbf{ a+b } &=& \mathbf{\sqrt{4ab+(a-b)^2}} \quad | \quad ab=45,\quad a-b = 12 \\ & a+b &=& \sqrt{4\cdot 45+12^2} \\ & a+b &=& \sqrt{324} \\ &\mathbf{ a+b} &=& \mathbf{18} \\ \hline \end{array} \)

The sum of the integers is 18

Beautiful- and a better diagram! Where did you make the diagram?

-\(tommarvoloriddle\)

​ Extremely Difficult German Problem

1.

\(\begin{array}{|rcll|} \hline \mathbf{\sin(\alpha)} &=& \mathbf{\dfrac{1}{x-7}} \qquad (1) \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(180^\circ-(90^\circ-\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(90^\circ+\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\cos(\alpha) }{x} \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\\\ \dfrac{ \sqrt{2}}{2(x-7)} &=& \dfrac{\cos(\alpha) }{x} \\\\ \mathbf{\cos(\alpha)} &=& \mathbf{ \dfrac{ x\sqrt{2}}{2(x-7)} } \qquad (2) \\ \hline \end{array} \)

Phythagoras:

\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)

\(\mathbf{x = 24\ cm}\)

What would happen after they break the policy?

THIS PROBLEM IS ALREADY ANSWERED BY HIGHLY QUALIFIED PEOPLE: JUST WANNA GET MY WAY IN(really similar though...)

Factors of -7 = ( 7 , -1 ) , ( -7 , 1 )

but since the sum is positive,

the positive has to be greater. 

So the 7 has to be larger than the 1 to make it positive.

So it is the first pair, ( 7 , -1 ) is correct.

Answer: 7 , -1

-\(tommarvoloriddle\)

Really off topic- wish there was a comment thing- for me, on my screen, the 2 length is greater than the 7 length.

You are supposed to pay? Where?

:) whooooops

um no. I was just showing my last steps, which was 15+3=18. Because 15 and 3 were a pair...

When you thought the answer is 2, this was your mistake:

You thought \(\Delta\) could be a two digit number, and assumed it was 12.

However, in the problem, it states "1\(\Delta\)". It cannot be "112"

So in the future, be careful, remember that when "1\(\Delta\)" happens, \(\Delta\) must be SINGLE digit number.

You read it wrong!!! Oops!

It said difference, not sum

This is my way, there might be a better way.

When it says "their product is 45" I just find its factor pairs

5 * 9

15 * 3

Those are the first to come into my mind

Then I instantly recognize 15 and 3 has a DIFFERENCE of 12.

You can make a system of equations, but that is too slow.

How do I pay?  I don’t see any links for making a payment.

Are you sure you know how to do this problem. You don’t seem like you know this kind of math.

Ok, rethinking this: for A)

try plugging number to find a pattern...

For b) you should be able to solve it with a simple knowledge of Sigma notation, try going to khan academy for review.

(maybe somebody else on the forum will help with a fully detailed answer.)

-deep regrets,

\(tommarvoloriddle\)

Wow.. That was better Phrased... Not my Question, But Thank You!

THX, tommarvoloriddle   !!!

Here's one other way

Let 2n be the smallest integer  and  2n + 18  the largest

The sum of these  divided by 2 will = the mean

So

[ 2n + 2n + 18 ]

____________   =     3            multiply both sides by 2

         2

2n + 2n +18  =  6

4n + 18  =  6          subtract  18 from both sides

4n  =  -12   divide both sides by 4

n =  - 3  

So...the smallest integer  = 2n  = 2(-3)  = - 6

The arithmetic mean of 10 consecutive even integers is 3. What is the least of these 10 even integers? If the mean is 3, then five of the numbers must be less than 3 and five greater than 3. The numbers less than 3 must be −6, −4, −2, 0, 2 and those greater than 3 must be 4, 6, 8, 10, and 12. The least of these is −6 .

-tommarvoloriddle

:)

-tommarvoloriddle

Welcome.....