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3rd route of 22.4 gives 2.8189

Therefore the cube would have sides 2.8189 decimeters, or 28.2 cm (3 significant figures)

Are you still looking for an answer? Please answer so I know if you need help or Not.

Heres an example:

Can any one help me find the mass density of a sphere the sphere has a diametor of 95mm. I need to convert my answer in to g/cm3 and kg/m3?

Density = Mass / Volume 
Volume = (4/3)πr³ = (4/3)π(95/2)³ = 448920.5 mm³ 

If z = 4+i3, then |z| = ?

\(\begin{array}{rcrcl} \boxed{~ z = a+ib \qquad |z| = \sqrt{a^2+b^2} ~} \end{array}\)

\(\begin{array}{rcrcl} z &=& 4+i\cdot 3 \\ |z| &=& \sqrt{4^2+3^2}\\ |z| &=& \sqrt{25}\\ |z| &=& 5 \end{array}\)

My day is not going well, I walk into school, and my friend says some VERY mean words to me. TThen shee said to me " Everyone's life would be better if you were dead", " People would be so happy if you were gone", Then she said " No one likes you, Your annoying,, way to happy, and your just ugly!!!" Usually that stuff doesn't affect me, but it is. :'(

0!

I'm like 99.9% Sure

i think: 

apatmen houses                         private houses 

         4                           :                       14

                                4/14=2/7

         2                                                    7

the answer:     2:7

\(\frac{4}{14} = \frac{2}{7}\)

Hey Chris, do I have to answer my own question :/  LOL

I wonder how easy it would be for either of us if the question were harder.

Say 3 had to be seperated and there were 25 others to go into the line as well.

How hard is this one?

It is too late for me to think about it now.  Maybe you might like to consider it though :)

This is what I reckon.

There are 25! ways the non trouble makers can be put in a line. 

Then there are 26 spots where the other 3 can be slipped in so that is    26P3 = 26*25*24 = 15600

So there would be  15600*25! ways.

And this is about 2.4*10^(29)   and that is a HUGE number of ways :))

Fri 15/1/16

1)  Jokes on the forum,  thanks Olivia3.14.... , Hayley and CPhill    

     http://web2.0calc.com/questions/math_75574

2)  The humour in this one appealed to me,  Thanks guest Hayley and Solveit  

     http://web2.0calc.com/questions/math_11311

3)  Chemistry        Thanks Nauseated and guest.

     http://web2.0calc.com/questions/chemistry-math_2

4)  Simultaneous equations - with a difference with a weird graph.  ://     Thanks guest and Melody.

     https://web2.0calc.com/questions/problem_76364

5)  Motion problem using formula OR calculus     

     http://web2.0calc.com/questions/65165#r5

6)  Co-ordinate geometry reflection.  Cool graph too     Thanks CPhill

     http://web2.0calc.com/questions/reflection-of-conic-with-line

7)  Finding zeros of a polynomial of degree 3   Thanks guest and CPhill

     http://web2.0calc.com/questions/math_85056

8)  Matrices  Thanks Heureka and CPhill

     http://web2.0calc.com/questions/matrix-pt-3

9)  Find the next terms in the sequence       Heureka's brilliance continues from yesterday :)

     http://web2.0calc.com/questions/sequence_3#r5

10)  A downloadable calculator with a DMS button.  Thanks guest - Much appreciated

                  (continued from yesterday)

      http://web2.0calc.com/questions/i-need-that-dms-button

@@ What is Happening?  [Wrap4]   Thurs 14/1/16   Sydney,  Australia Time 11:07 pm   ♪ ♫

Hi everyone,

Great answers today were provided by Solveit, CPhill, Hayley1, Admin, calmmylife, Heureka, gibsonj338, xvxvxv and GenericUserNameHere.    Thank you      

Interest Posts: 

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.

1)  How long will it take.  I like these problems.  Thanks Solviet, Hayley, Melody Guest and CPhill.

how to solve what is the sum of the first ten terms of the geometric sequence 4,8,16,.....?

\(\begin{array}{rcl} a_1 &=& 2^2 \\ a_2 &=& 2^3 \\ a_3 &=& 2^4 \\ \cdots \\ a_n &=& 2^{n+1} \end{array}\)

 \(\begin{array}{rcrcl} s_n &=& a_1 &+& a_2+a_3+a_4+ \cdots + a_{n-1} + a_n \\ \hline s_n &=& 2^2 &+& 2^3+2^4+2^5+ \cdots + 2^n + 2^{n+1} \\ 2\cdot s_n &=& && 2^3+2^4+2^5+ \cdots + 2^n+2^{n+1} + 2^{n+2} \\ \hline s_n - 2\cdot s_n &=& 2^2 - 2^{n+2} \\ s_n(1 - 2) &=& 2^2 - 2^{n}\cdot 2^2 \\ -s_n &=& 2^2 - 2^{n}\cdot 2^2 \\ s_n &=& 2^{n}\cdot 2^2 - 2^2\\ s_n &=& 2^2\cdot (2^{n}-1)\\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{4\cdot (2^{n}-1)} \\\\ s_{10} &=& 4\cdot (2^{10}-1)\\ s_{10} &=& 4\cdot (1024-1)\\ s_{10} &=& 4\cdot 1023\\ \mathbf{s_{10}} &\mathbf{=} & \mathbf{4092} \end{array}\)

The sum of the first ten terms of the geometric sequence is 4092

127,28188282828282828828282

whats the angle for C if a = 8 b = 6 c = 7

cos-rule

\(\begin{array}{rcl} c^2 &=& a^2+b^2-2ab\cos{(C)} \\ c^2+2ab\cos{(C)} &=& a^2+b^2 \\ 2ab\cos{(C)} &=& a^2+b^2-c^2 \\ \cos{(C)} &=& \frac{a^2+b^2-c^2} {2ab} \\ \cos{(C)} &=& \frac{8^2+6^2-7^2} {2\cdot 8 \cdot 6} \\ \cos{(C)} &=& \frac{64+36-49} {96} \\ \cos{(C)} &=& \frac{51} {96} = \frac{17} {32} \\ \cos{(C)} &=& 0.53125 \\ C &=& 57.9100487437^\circ \end{array}\)

4 people take 2.5 hours to do 1/2 the job

people 4-10  4*5/2=10       job 1/2-1/3 1/2*2/3=1/3

10 people take x hours to do 1/3 the job

 so: (150*2*2)/(3*5)=40

i did the same :P :D

I looked and could not find one either ://

That is really strange I think.......

Maybe someone else can find one???

There are a few sites like this one though

It's just 1.

89/1416 = 0.0628531073446328

as a %, = 6.285%

A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 0.5 × 10ȳ seconds, what is the value of impuls

I assume

\(F = 520\ N\)

\(t = 0.5 \cdot 10^{-3} \ s\)

\(p = \text{impuls in}\ Ns\)

\(\begin{array}{rcl} p &=& F\cdot t\\ p &=& 520\ N \cdot 0.5 \cdot 10^{-3} \ s\\ p &=& 520\cdot 0.5 \cdot 10^{-3} \ Ns\\ p &=& 520\cdot 0.0005 \ Ns\\ p &=& 0.26 \ Ns\\ \end{array}\)

Nauseated, since you are such a 'genius'  [interprete as 'know-it-all' ] why don't you just answer the question!!

Perhaps this would present a problem for you ? 

Thanks Heureka,

Here is a site that you will like guest :)

It is the

Online Encyclopeadia of Integer Sequences  :)

how do you get rid of the bottom? [(x^3-2x)-(a^3-2a)]/(x-a)

\(\begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-2x-a^3+2a } {x-a}\\ &=& \frac{ x^3-a^3-2x+2a } {x-a}\\ &=& \frac{ x^3-a^3-2(x-a) } {x-a}\\ &=& \frac{ x^3-a^3 }{x-a} - 2\cdot( \frac{ x-a } {x-a} ) \\ &=& \frac{ x^3-a^3 }{x-a} - 2\\ \end{array}\\ \boxed{~ \text{Difference of two cubes:}\\ \begin{array}{rcl} x^3-a^3 &=& (x-a)(x^2+x\cdot a+a^2) \end{array} ~}\\ \begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-a^3 }{x-a} - 2\\ &=&\frac{ (x-a)(x^2+x\cdot a+a^2) }{x-a} - 2\\ &=& x^2+x\cdot a+a^2 - 2\\ \end{array} \)

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

We need the prime number table

\(\text{The prime numbers. }\\ \begin{array}{|r|r||r|r|||r|r|||r|r|||r|r||r|r|} \hline n & p(n) & n & p(n) & n & p(n)& n & p(n)& n & p(n)& n & p(n) \\ \hline 1 & 2 &21 & 73 &41 & 179 & 61 & 283 &81 & 419 & 101 & 547 \\ 2 & 3 &22 & 79 &42 & 181 & 62 & 293 &82 & 421 & \cdots & \cdots \\ 3 & 5 &23 & 83 &43 & 191 & 63 & 307 &83 & 431 & \cdots & \cdots \\ 4 & 7 &24 & 89 &44 & 193 & 64 & 311 &84 & 433 & \cdots & \cdots \\ 5 & 11 &25 & 97 &45 & 197 & 65 & 313 &85 & 439 & \cdots & \cdots \\ 6 & 13 &26 & 101 &46 & 199 & 66 & 317 &86 & 443 & \cdots & \cdots \\ 7 & 17&27 & 103 &47 & 211 & 67 & 331 &87 & 449 & \cdots & \cdots \\ 8 & 19 &28 & 107 &48 & 223 & 68 & 337 &88 & 457 & \cdots & \cdots \\ 9 & 23 &29 & 109 &49 & 227 & 69 & 347 &89 & 461 & \cdots & \cdots \\ 10 & 29&30 & 113 &50 & 229 & 70 & 349 &90 & 463 & \cdots & \cdots \\ 11 & 31& 31 & 127& 51 & 233& 71 & 353 &91 & 467 & \cdots & \cdots \\ 12 & 37&32 & 131 &52 & 239 & 72 & 359 &92 & 479 & \cdots & \cdots \\ 13 & 41& 33 & 137 &53 & 241& 73 & 367 &93 & 487 & \cdots & \cdots \\ 14 & 43& 34 & 139 &54 & 251& 74 & 373 &94 & 491 & \cdots & \cdots \\ 15 & 47& 35 & 149 &55 & 257& 75 & 379 &95 & 499 & \cdots & \cdots \\ 16 & 53& 36 & 151 &56 & 263& 76 & 383 &96 & 503 & \cdots & \cdots \\ 17 & 59& 37 & 157 &57 & 269& 77 & 389 &97 & 509 & \cdots & \cdots \\ 18 & 61& 38 & 163 &58 & 271& 78 & 397 &98 & 521 & \cdots & \cdots \\ 19 & 67& 39 & 167 &59 & 277& 79 & 401 &99 & 523 & \cdots & \cdots \\ 20 & 71& 40 & 173 &60 & 281& 80 & 409 &100 & 541 & \cdots & \cdots \\ \hline \end{array} \)

The formula is \(\boxed{~ a_n = a_{n-1} + p(n) -p(n-1) +1 \qquad n \ge 2 \qquad a_1 = 2 ~}\)

\(\begin{array}{lcl} a_1 &=& 2 \\ \hline a_2 &=& a_{1} + p(2) -p(1) +1 \\ &=& 2 +3-2+1 \\ &=& 4 \\ \hline a_3 &=& a_{2} + p(3) -p(2) +1 \\ &=& 4 +5-3+1 \\ &=& 7 \\ \hline a_4 &=& a_{3} + p(4) -p(3) +1 \\ &=& 7 +7-5+1 \\ &=& 10 \\ \hline a_5 &=& a_{4} + p(5) -p(4) +1 \\ &=& 10 +11-7+1 \\ &=& 15 \\ \hline a_6 &=& a_{5} + p(6) -p(5) +1 \\ &=& 15 +13-11+1 \\ &=& 18 \\ \hline a_7 &=& a_{6} + p(7) -p(6) +1 \\ &=& 18 +17-13+1 \\ &=& 23 \\ \hline a_8 &=& a_{7} + p(8) -p(7) +1 \\ &=& 23 +19-17+1 \\ &=& 26 \\ \hline \hline a_9 &=& a_{8} + p(9) -p(8) +1 \\ &=& 26 +23-19+1 \\ &=& 31 \\ \hline a_{10} &=& a_{9} + p(10) -p(9) +1 \\ &=& 31 +29-23+1 \\ &=& 38 \\ \hline a_{11} &=& a_{10} + p(11) -p(10) +1 \\ &=& 38 +31-29+1 \\ &=& 41 \\ \hline a_{12} &=& a_{11} + p(12) -p(11) +1 \\ &=& 41 +37-31+1 \\ &=& 48 \\ \hline \cdots \\ \hline a_{100} &=& a_{99} + p(100) -p(99) +1 \\ &=& 621 +541-523+1 \\ &=& 640 \\ \hline \end{array} \)

\(a_1 \cdots a_{50}\\ \begin{array}{|r|r||r|r|||r|r|||r|r|||r|r|} \hline n & a(n) & n & a(n) & n & a(n)& n & a(n)& n & a(n)\\ \hline 1 & 2 & 11 & 41 & 21 & 93 &31 & 157 &41 & 219\\ 2 & 4 & 12 & 48 & 22 & 100 &32 & 162&42 & 222\\ 3 & 7 & 13 & 53 & 23 & 105 &33 & 169&43 & 233\\ 4 & 10 & 14 & 56 & 24 & 112 &34 & 172&44 & 236\\ 5 & 15 & 15 & 61 & 25 & 121 &35 & 183&45 & 241\\ 6 & 18 & 16 & 68 & 26 & 126 &36 & 186&46 & 244\\ 7 & 23 & 17 & 75 & 27 & 129 &37 & 193&47 & 257\\ 8 & 26 & 18 & 78 & 28 & 134 &38 & 200&48 & 270\\ 9 & 31 & 19 & 85 & 29 & 137 &39 & 205&49 & 275\\ 10 & 38 & 20 & 90 & 30 & 142 &40 & 212&50 & 278\\ \hline \end{array} \)

\(a_{51} \cdots a_{101}\\ \begin{array}{|r|r||r|r|||r|r|||r|r|||r|r||r|r|} \hline n & a(n) & n & a(n) & n & a(n)& n & a(n)& n & a(n)& n & a(n) \\ \hline 51 & 283 &61 & 343&71 & 423&81 & 499&91 & 557&101 & 647 \\ 52 & 290 &62 & 354&72 & 430&82 & 502&92 & 570&\cdots & \cdots \\ 53 & 293 &63 & 369&73 & 439&83 & 513&93 & 579&\cdots & \cdots \\ 54 & 304 &64 & 374&74 & 446&84 & 516&94 & 584&\cdots & \cdots \\ 55 & 311 &65 & 377&75 & 453&85 & 523&95 & 593&\cdots & \cdots \\ 56 & 318 &66 & 382&76 & 458&86 & 528&96 & 598&\cdots & \cdots \\ 57 & 325 &67 & 397&77 & 465&87 & 535&97 & 605&\cdots & \cdots \\ 58 & 328 &68 & 404&78 & 474&88 & 544&98 & 618&\cdots & \cdots \\ 59 & 335 &69 & 415&79 & 479&89 & 549&99 & 621&\cdots & \cdots \\ 60 & 340 &70 & 418&80 & 488&90 & 552&100 & 640&\cdots & \cdots \\ \hline \end{array} \)

I'll probably do it exactly the same as Chris, I just want to play with it.  :)

4 people take 2.5 hours to do 1/2 the job

so

4 people will take 5 hours to do the whole job

so

1 person will take 20 hours to do the whole job

so

10 people will take 2 hours to do the whole job

so

10 people will take 2/3 hours to do 1/3 of the job

2/3 of an hour is 40 minutes.     So it will take 10 people 40minutes to do 1/3 of the job.     Ta~da