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Did you see my answer over here Nickolas?

Guest asker,

Since you are too lazy to delete the dollar signs and make your question readable I do not think you should bother enrolling in engineering.

I do not want to enter any building you design.  No Earthquake will be needed to induce that disaster.

Very impressive Heureka :)

10^n = sqrt(10^46/0.0001)/10^6

        = sqrt(10^50)  / 10^6

        = 10^25  / 10^6

        = 10^19                     n = 19

L = 25 T^4 /  H^2

L = 25 (4^4) / (8^2)

L = 25 *4^4/(8^2) = 100        I do not know what the units are....You should be able to do these problems...all of the variables are given...just plug and play.

Solve for n:
10^n = sqrt(10^46/0.0001)/10^6

sqrt(10^46/0.0001)/10^6 = sqrt(10^46/(1/10000))/1000000:
10^n = sqrt(10^46/(1/10000))/1000000

10000000000000000000 = 2^19·5^19:
2^n·5^n = 2^19·5^19

Equate exponents of 2 and 5 on both sides:
n = 19 and n = 19

All equations give n = 19 as the solution:
n = 19

Thankyou for explaing that KeylimePi 

I see 

Thanks for your help guys yes I was alittle confused with the last post Ep did for this same question I didnt really no what to do at the part were 1/4 of the elvation of city (C)! 

I wasint sure for 1/4 part. 

Thanks Melody 

ElectricPavlov

KeylimePie 
 

An infinite geometric series has first term -3/2 and sums to twice the common ratio.

Find the sum of all possible values for the common ratio.

\(\text{Let $a_1=a=-\dfrac{3}{2} $} \\ \text{Let the common ratio $=r$} \\ \text{Let the sum of the infinite geometric series $= \dfrac{a}{1-r} $ } \)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \mathbf{2r} \\\\ a &=& 2r(1-r) \\ a &=& 2r-2r^2 \\ 2r^2-2r+a &=& 0 \quad &|\quad a = -\dfrac{3}{2} \\ 2r^2-2r-\dfrac{3}{2} &=& 0 \quad &|\quad \cdot 2 \\ 4r^2-4r-3 &=& 0 \\\\ r &=& \dfrac{4\pm \sqrt{4^2-4\cdot 4\cdot (-3)} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2+4^2\cdot 3} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2\cdot 4} }{2\cdot 4} \\ &=& \dfrac{4\pm 4\cdot 2 }{2\cdot 4} \\ &=& \dfrac{1\pm 2 }{2} \\\\ r_1 &=& \dfrac{1+ 2 }{2} \\ \mathbf{ r_1 } &=& \mathbf{ \dfrac{3}{2} } \\\\ r_2 &=& \dfrac{1- 2 }{2} \\ \mathbf{ r_2 } &=& \mathbf{ -\dfrac{1}{2} } \\ \hline \end{array} \)

The sum of all possible values for the common ratio \(\dfrac{3}{2}-\dfrac{1}{2} = \mathbf{1}\)

Four positive integers \(w\), \(x\), \(y\),and \(z\) satisfy \(wx + w + x = 524\), \(xy + x + y = 146\), \(yz + y + z = 104\), and \(wxyz= 8!\).
What are \(w\), \(x\), \(y\), and \(z\) ?

Solution by substitution:

\(\begin{array}{|rcll|} \hline wx + w + x &=& 524 \\ x(w+1)+w &=& 524 \\ x(w+1) &=& 524-w \\ \mathbf{x} &=& \mathbf{\dfrac{524-w} {w+1}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xy + x + y &=& 146 \\ y(x+1) + x &=& 146 \\ y(x+1) &=& 146-x \\ \mathbf{y} &=& \mathbf{\dfrac{146-x} {x+1}} \\\\ y &=& \dfrac{146-\dfrac{524-w} {w+1}} {\dfrac{524-w} {w+1}+1} \\ \ldots \\ \mathbf{y} &=& \mathbf{\dfrac{147w-378}{525} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline yz + y + z &=& 104 \\ z(y+1) + x &=& 146 \\ z(y+1) &=& 104-y \\ \mathbf{z} &=& \mathbf{\dfrac{104-y} {y+1}} \\ z &=& \dfrac{104-\dfrac{147w-378}{525}} {\dfrac{147w-378}{525}+1} \\ \ldots \\ \mathbf{z} &=& \mathbf{\dfrac{54978-147w}{147(w+1)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{wxyz} &=& \mathbf{8!} \\ wxyz &=& 40320 \\ 40320 &=& w \left(\dfrac{524-w} {w+1}\right) \left(\dfrac{147w-378}{525} \right) \left(\dfrac{54978-147w}{147(w+1)} \right) \\ 40320 &=& \dfrac{w(524-w)(147w-378)(54978-147w)} {77175(w+1)^2} \\ 3111696000 (w+1)^2 &=& w(524-w)(147w-378 )(54978-147w ) \\ \ldots \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline & -21609 w^4 + 19460448 w^3 - 1173047652 w^2 + 17112994416 w + 3111696000 &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Solutions: }\\ &\mathbf{ w } &=& \mathbf{24}\ \checkmark \qquad \text{integer} \\ & w&=&0.17961 \\ &w&=&39.918 \\ &w&=&36.83 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{524-w} {w+1}} \\ x &=& \dfrac{524-24} {24+1} \\ x &=& \dfrac{500} {25} \\ \mathbf{x} &=& \mathbf{20} \\\\ \mathbf{y} &=& \mathbf{\dfrac{146-x} {x+1}} \\ y &=& \dfrac{146-20} {20+1} \\ y &=& \dfrac{126} {21} \\ \mathbf{y} &=& \mathbf{6} \\\\ \mathbf{z} &=& \mathbf{\dfrac{104-y} {y+1}} \\ z &=& \dfrac{104-6} {6+1} \\ z &=& \dfrac{98} {7} \\ \mathbf{z} &=& \mathbf{14} \\ \hline \end{array}\)

Until this gets fixed, you should be able to use another service to upload your image.

Given the data you supply, your calculations are correct.

Yah.....Happens to me too    ~EP

Wait, so technically the answer is 0?

It was not EPs post. EP has assured me he did not write it.

For the benefit of others,  EP and Nickolas are both being trolled.    

Nickolas, try drawing a picture of the sea. And put the things you know at their correct depth.

You are told A and C you can put them in.  Then you can think about D

Like this only it is not long enough.

LOL hahahahahah! Good one EP!

You should put EP’s post back. It’s the best post in the past week! I posted a copy on #2 

You should put your post back too!  It’s not as funny but it’s true!

Thanks for your great answer keylime.

I would like to suggest that you do not use sentences like 

"This wasn't a very challenging problem. "

What is easy for you may well be very difficult for someone else.