All Answers 357301 Answers

What is your game?  I know BM is one of your puppets, TPM (The Puppet Master), but I don’t yet know what your game is. Is it some kind of self-styled therapy to deal with your hyperactive, chaotic brain? 

If so, please continue. Trolling you and the Blarney Banker is my self-styled cathartic therapy:  like squeezing zits, lancing boils, and breaking dams of batshit-stupid that keep the offal from flowing through the sewers of mystical Camelot.   

BM This is an example of your disconnect to reality: Compare Sir CPhill’s answer to your meandering pile of incomprehensible slop. 

CPhill’s answer is compact, to the point, and clear. His comment is relevant and concise.

Your answer meanders all over Hades. By the time anyone wades though and digests this hog slop of discontinuity, the reader forgets the question, as the toxic brew induces a mental fog and accelerates brain atrophy.

An ideal gas is connected in a closed cylinder in a moving piston the initial pressure is 1ATM and the start is 1 L. Constant pressure gas is heated up to double its volume then heated to constant volume to double the pressure and finally expands adiabatic until the temperature drops to its initial value. 

un gas ideal esta conectado en un cilindro cerrado en un piston movil la presion inicial es 1atm y el inicio es 1L. se calienta el gas a presion constante hasta ducplicar su volumen despues se calienta a volmen constante hasta duplicar la presion y finalmente se expande adiabaticamente hasta que la temperatura deciende a su valor inical

Hallo dimcorak!

0. Ein ideales Gas ist in einem Zylinder mit einem beweglichen Kolben eingeschlossen.

    Das Anfangsvolumen ist 1 Liter, der Anfangsdruck 1013 hPa.

1. Das Volumen wird isobar verdoppelt.

2. Der Druck wird isochor verdoppelt.

3. Die Temperatur sinkt bei adiabatischer Expansion auf die Anfangstemperatur.

(Only for my understanding.)

$Pressure, temperature\ and\ volume\\ according\ to\ the\ three\ state\ changes.$

0.

 $p_0=1013\ hPa\\ V_0= 1000\ cm^3$

T is undetermined. I suggest:

$T_0=20°C=293K$

1. isobar

$p_1=1013\ hPa\\ V_1= 2000\ cm^3$

$\frac{V_1}{T_1}=\frac{V_0}{T_0}\\ T_1=\frac{V_1\cdot T_0}{V_0}=\frac{2l\cdot293K}{1l}\\ \color{blue} T_1=586K$

2. isochor

$\color{blue}V_2=V_1=2000\ cm^3\\ \color{blue}p_2=2p_1=2026hPa\\ \frac{p_1}{T_1}=\frac{p_2}{T_2}\\ T_2=\frac{p_2T_1}{p_1}=\frac{2026\cdot 586K}{1013}\\ \color{blue}T_2=1172K$

3. addiabat, isentrop

$\kappa_{gas\ ideal}=1.4$

$\color{blue }T_3=T_0=293K$

$\frac{T_2}{T_0}=(\frac{V_2}{V_3})^{\kappa-1}\\ (\frac{T_0}{T_2})^{\frac{1}{\kappa-1}}=\frac{V_3}{V_2}\\ V_3=V_2(\frac{T_0}{T_2})^{\frac{1}{\kappa-1}}$ 

$V_3=2000cm^3(\frac{1172K}{293K})^{\frac{1}{1.4-1}}$$\frac{p_0}{p_2}=(\frac{V_2}{V_3})^{\kappa}$

$V_3=64000\ cm^3$

$\frac{p_3}{p_2}=(\frac{V_2}{V_3})^{\kappa}\\ p_3=p_2(\frac{V_2}{V_3})^{\kappa}\\ p_3=2026hPa(\frac{2000cm^3}{64000cm^3})^{1.4}\\$

$p_3=15.828\ hPa$

I hope I could help you.
If you have any questions, please contact us!

greeting

  !

A mistake ?

Life's just the way it is...

I think heureka misunderstood the question !! I believe the questioner means the entire number as one large number. It is separated in sixes because it is easier to read. If that is correct, then the entire number is composite according to Rabin-Miller Test. It has 2 large prime factors:

= 77777 733222 232222 222222 222223 × 110000 013000 001700 000190 000023

A polynomial has exactly 7 roots. The graph tends towards positive infinity in both directions. Write down some information about what you know about his polynomial.

MMm 

Well the leading coefficient is positive.

The degree is even and great or equal to 8.

That is about all I have I guess.

I was thinking there are exactly 7 real roots (plus some imaginary ones) but maybe you mean exactly 7 roots which includes real and imaginary.

If that is what you mean then my answer is not correct.

If it has exactly 7 roots then one of the roots must be a double.

It will have a degree of 8 (or more if there are more double roots, degree must be even )

Definitely a positve leading coefficient.

That is my 2 bob worth! :)

Yep, but thanx anyways Chris!...

Never mind, I figured it out.

1. $\frac{\frac{7}{4}}{10}=\frac{\frac{x}{y}}{30}$

2. $\frac{3}{3}*\frac{7}{4}=\frac{x}{4}$

3. $\frac{21}{12}=\frac{x}{y}$

Finally: $x=21; y=12$

Solve algebraically and graphically !

Thanks Chris but that is exactly what guest and I already did.

I guess it must be 3 times lucky!

Thanks Chris,

Silly me, I did not see the doubling his money bit of the question!

Of course my answer has to be doubled to get that profit! 

What is the wavelength of a wave having a frequency of 3.76 multiplied by 10^14 s^-1?

(please go show ALL work to get to the answer)

$\begin{array}{llll} \text{Here is the formula of the wavelength } (\lambda) \\ \text{depending on } v (\text{velocity of the wave }) \text{ and } f (\text{frequency}). \\ \end{array}$

$\begin{array}{rrr} && \mathbf{\lambda = \dfrac{v}{f} } \\ \end{array}$

$\begin{array}{llll} \mathbf{\lambda} & \text{ (Lambda) is the wavelength } \\ \mathbf{v} & \text{ is the velocity of the wave (default is velocity of light in vacuum: } 299792458 \frac{m}{s} ) \\ \mathbf{f} & \text{ is the frequency } \\ \end{array}$

$\begin{array}{|rcll|} \hline \lambda &=& \dfrac{v}{f} \quad & | \quad v=299792458 \frac{m}{s} \qquad f = 3.76\ \cdot 10^{14} \frac{1}{s} \\\\ &=& \dfrac{299792458 \frac{m}{s}} {3.76\ \cdot 10^{14} \frac{1}{s}} \\\\ &=& \dfrac{299792458 } {3.76\ \cdot 10^{14} } \cdot \frac{m}{s} \cdot \frac{s}{1} \\\\ &=& \dfrac{299792458 } {3.76\ \cdot 10^{14} } \ m \\\\ &=& \dfrac{2997.92458\cdot 10^{5} } {3.76\ \cdot 10^{14} } \ m \\\\ &=& \dfrac{2997.92458} {3.76}\cdot 10^{5-14} \ m \\\\ &=& \dfrac{2997.92458} {3.76}\cdot 10^{-9} \ m \quad & | \quad nm = 10^{-9} \ m \\\\ &=& \dfrac{2997.92458} {3.76}\ nm \\\\ &\mathbf{=}& \mathbf{797.320367021 \ nm} \\ \hline \end{array}$

The wavelength of a wave is 797 nm

Is this number prime or composite? If not prime, can you find its prime factors? Thank you.

8555 551665 556208 555624 588899 297778 663111 342433 333481 111129

$\begin{array}{|rcll|} \hline 8555 &=& 5 * 29 * 59 \\ 551665 &=& 5 * 19 * 5807 \\ 556208 &=& 2^4 * 34763 \\ 555624 &=& 2^3 * 3^2 * 7717 \\ 588899 &=& 127 * 4637 \\ 297778 &=& 2 * 13^2 * 881 \\ 663111 &=& 3^2 * 73679 \\ 342433 &=& 7 * 13 * 53 * 71 \\ 333481 &=& 37 * 9013 \\ 111129 &=& 3 * 17 * 2179 \\ \hline \end{array}$

[ 55t ] / [ t^2 + 25 ] ≥ 5     multiply both sides by  t^2 + 25

55t  ≥ 5 [ t^2 + 25]       simplify

55t ≥ 5t^2  + `125     divide  through by 5

11t ≥ t^2 + 25       subtract  11t  from both sides

0 ≥ t^2 - 11t + 25   and we can write

t^2 - 11t + 25 ≤ 0

Let's solve this

t^2 - 11t + 25 = 0

Putting this into the quadratic formula......we have the [approximate] solutions

t ≈ 3.2  hrs    and    t ≈  7.8  hrs

Choosing a test value between these two values  [ I'll choose 4 ]  and putting this into

t^2 - 11t + 25 ≤ 0  →  4^2 - 11(4) + 25  =  16 - 44 + 25  = -3  

So....since 4 makes the inequality true.....the interval between 3.2 hrs  and 7.8 hrs is the correct solution.....i.e, the concentration will be ≥ 5  between these two times

Here's the graph, substituting x for t : https://www.desmos.com/calculator/cjgoqwvrj5

Notice that  [ 55t ] / [ t^2 + 25 ]  is ≤ 5  between these two times

Note......20  =   2 x 10¹  

So....we have

[2.56 x 10⁵ ] / [ 2 x 10¹ ]  =

[ 2.56 / 2 ]  x [10⁵ / 10¹ ]  =

[ 1.28 ] x  [ 10⁴ ]  =  

12800  km^3  = volume of  Lake Rushy

Multiplying through by -2  sets it up for completing the square........

once you got -q^2/2 + 2q - 10 = 0, why would you wanna multiply both sides by -2? is a denominator not allowed? 

EVOLUTION !!.

Solve for q:
sqrt(q^2/2 + 11) = q - 1

Raise both sides to the power of two:
q^2/2 + 11 = (q - 1)^2

Expand out terms of the right hand side:
q^2/2 + 11 = q^2 - 2 q + 1

Subtract q^2 - 2 q + 1 from both sides:
-q^2/2 + 2 q + 10 = 0

Multiply both sides by -2:
q^2 - 4 q - 20 = 0

Add 20 to both sides:
q^2 - 4 q = 20

Add 4 to both sides:
q^2 - 4 q + 4 = 24

Write the left hand side as a square:
(q - 2)^2 = 24

Take the square root of both sides:
q - 2 = 2 sqrt(6) or q - 2 = -2 sqrt(6)

Add 2 to both sides:
q = 2 + 2 sqrt(6) or q - 2 = -2 sqrt(6)

Add 2 to both sides:
q = 2 + 2 sqrt(6) or q = 2 - 2 sqrt(6)

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 - 2 sqrt(6))^2 + 11) = sqrt(25 - 4 sqrt(6)) ≈ 3.89898
q - 1 ⇒ (2 - 2 sqrt(6)) - 1 = 1 - 2 sqrt(6) ≈ -3.89898:
So this solution is incorrect

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 + 2 sqrt(6))^2 + 11) = sqrt(25 + 4 sqrt(6)) ≈ 5.89898
q - 1 ⇒ (2 + 2 sqrt(6)) - 1 = 1 + 2 sqrt(6) ≈ 5.89898:
So this solution is correct

The solution is: q = 2 + 2 sqrt(6)

Yea, "Mind Reader Psycho !!".

Why are you obsessed with large numbers?

1.5241578775 1564791571 4744640673 6584300407 4106033398 9936102274 6700769660 7026298183 3538950468 8864989143 5630676833 6462276165 3986049765 0828260258 9615170273 2973042434 3528139368 3906723701 6735057835 7224508852 0444718482 6428750208 9457981404 6906149443 6146991699 4554245545 9501648356 8362937505 8511850366 7322124313 2167496719 0421704957 9458486403 6651578429 9374718410 5001643719 0906091729 44618081 E+382.

−3√ ​180 ​ ​​ + 4√ ​45 ​ ​​ =

-3 √ [ 4 * 45 ]  + 4 √45  =

-3 √4*√45 + 4√45  =

-3 *2 *√45 + 4√45 =

-6√45 + 4√45  =

-2√45  =

-2* √9 * √5  =

-2 * 3 * √5  =

-6√5