All Answers 356018 Answers

Key in  →      ( 50,031,545,098,999,707) ^ (1 / 35)

[ You should get "3" as your answer  ]

(64/729)^x = 243/32

(64/ 729)^x  =  3^5  / 2^5

( 2^6 / 3^6 )^x  = ( 2^5 / 3^5) ^(-1)

2^(6x)  / 3^(6x)  =  2^(-5) / 3^(-5)

(2/3)^(6x)  = (2/3)^(-5)    →  6x  = -5   →  x  = - 5 / 6

MathNerd I think that is incorrect. You could choose the 2 people that each of differnt foods in 4C2 which is 6 so it is 6 * 1/81 = 6/81 = 2/27

Make a noose².

Let   < YTU  = x

Let   < XUT  = y

Let   < TVU  = z

< YIT  = <XIU  = 71

And we have the following system

(1/2) x + (1/2)y  + 109  = 180    →   x + y  =  142

x + y + z  = 180

(1/2)x + 71  + (1/2)y + 71 + 109 + z  = 360  →    x + y + 2z  = 218

So....

x + y  =  142      (1)

x + y + z  = 180   (2)

x + y + 2z  = 218  (3)

Subtract  (1)  from (2)  and z = 38° = < TVU  = "angle V "

And.....  y  = 142 – x  →   (1/2)y  = 71 - (1/2)x

As Melody  always says, "This is what I think. "

Hint:  13 is a factor of both those :)

The one-way commuting times from home to work for all employees working at a large company have a bell-shaped curve with a mean of 31 minutes and a standard deviation of 8 minutes. Using the empirical rule, find the approximate percentages of the employees at this company who have one-way commuting times in the following intervals. 

what's the answer 

"Guest's "  answer isn't unique.....for example......

117649  = 343^2

117649 =    49^3

262144  = 512^2

262144  = 64^3

I suspect that, integer-wise, this may be all

[ 6 * 6 ] ^3   = [36]^3   = 46.656  ..... too small

[ 7 * 7 ] ^3 =   [49]^3  = 117649

[8 * 8 ] ^3  =   [64]^3  = 262144

[9 * 9 ] ^3  =   [81]^3  =  531441

[ 10 * 10] ^3  = [100]^3  = 1,000,000  ...... too large

P = 9^x

3^(2x) + 3 → (3^2)^x + 3 → 9^x + 3 → P + 3

3^(2x+3) →  (3^(2x))*3^3 → (9^x)*27 → 27P

.

Step by Step :

1. On segment AC, construct circle with center at A and radius AC

2. And construct circle with center at C and radius AC

3. These circles will intersect at B and  D

4. And AB = BC = AC  = CD = AD  since they are equal radii

5.  Now, ABCD forms a rhombus.....and, by definition of a rhombus, its diagonals AC and BD are  perpendicular bisectors of each other.

Hope that helps !!!

Thanks Alna, for your answer! Yes, I'm sorry for the mistake, I just didn't notice it, it's actually :h=-T(dv/dT) at constant p.

Here's my answer for this question: 

\(H=U+pv\)(H is the enthalpy)

\(H=Q+W+pv\)(U=Q+W)

\(dH=dQ+dW+d(pv)\)

knowing that \(dQ={C}_{v}dT+hdp\)

\(dH={C}_{v}dT+hdp-pdv+pdv+vdp\)

\(dH={C}_{v}dT+hdp+vdp\)

\(dH={C}_{v}dT+(h+v)dp\)

H is a state function, that means it's a total  differential:

\(\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T}=\begin{pmatrix} \frac{\partial({h+v})} {\partial T} \end{pmatrix}_{p} \)

\(\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} =\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} \)(1)

On the other hand we have:

\(dQ={C}_{v}dT+hdp\)\(*(\frac{1}{T})\)

\(dS=\frac{{C}_{v}}{T}dT+\frac{h}{T}dp\)

S is a state function, that means it's a total  differential:

\(\begin{pmatrix} \frac{\partial({C}_{v}/T)} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial({h/T})} {\partial T} \end{pmatrix}_{p}\)

\(\frac{1}{T}\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \frac{ {T} \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T²}}{T²}\)

\(\begin{pmatrix} \frac{\partial{C}_{v}} {\partial p} \end{pmatrix}_{T} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} -\frac{h}{T}\)(2)

(1)=(2)

\(\begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} + \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = \begin{pmatrix} \frac{\partial{h}} {\partial T} \end{pmatrix}_{p} - \frac{h}{T}\)

\( \begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} = - \frac{h}{T}\)

\(h= -T\begin{pmatrix} \frac{\partial{v}} {\partial T} \end{pmatrix}_{p} \)

I believe that the answer is   11C2  = n ( n -1) / 2   =  (11)(10)  / 2   =  55

Each point will have 10 lines connecting it to every other point, but we must divide by 2 to  prevent "double couning" since, for example, a line through point A and point E is the same as a line  through point E and point A

2x+11=-19-3x |+3x

5x+11=-19      |-11

      5x=-30      | :5

       x=-6

2x+11=-19-3x     First let's add 3x to both sides to get

5x +11 = -19       then subtract 11 from both sides

5x = -30               then divide both sides by 5

x = -6

P= 9^x

Log P = x log 9

Log P/Log 9 = x

1.04795 Log P = x      SUb this into the second equation

3^(2x) +3

3^(2(1.04795 LogP)) +3

3^(2.0959 Log P)  + 3 

That looks like the equation for the position of the particle (ball or rock or whatever)

d/dt of that will give you the value of the velocity at time t:

d /dt = -32t + 100 = v at time t

Possible derivation:
d/dx(sqrt(x/(1 + x^2)))
Using the chain rule, d/dx(sqrt(x/(x^2 + 1))) = ( dsqrt(u))/( du) 0, where u = x/(x^2 + 1) and ( d)/( du)(sqrt(u)) = 1/(2 sqrt(u)):
 = (d/dx(x/(1 + x^2)))/(2 sqrt(x/(1 + x^2)))
Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = x and v = x^2 + 1:
 = ((x^2 + 1) d/dx(x) - x d/dx(1 + x^2))/((x^2 + 1)^2) 1/(2 sqrt(x/(1 + x^2)))
The derivative of x is 1:
 = (-x (d/dx(1 + x^2)) + 1 (1 + x^2))/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
Simplify the expression:
 = (1 + x^2 - x (d/dx(1 + x^2)))/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
Differentiate the sum term by term:
 = (1 + x^2 - d/dx(1) + d/dx(x^2) x)/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
The derivative of 1 is zero:
 = (1 + x^2 - x (d/dx(x^2) + 0))/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
Simplify the expression:
 = (1 + x^2 - x (d/dx(x^2)))/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2: d/dx(x^2) = 2 x:
 = (1 + x^2 - 2 x x)/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)
Simplify the expression:
Answer: |= (1 - x^2)/(2 sqrt(x/(1 + x^2)) (1 + x^2)^2)

OK, young person. I got your number:

531,441 =729^2

531,441 =81^3

1,000,000 =1,000^2

1,000,000 =100^3

I know you said "6-digits long", but this is 7-digits long!.

Take the integral:
 integral u^2 sqrt(a^2 - u^2) du
For the integrand u^2 sqrt(a^2 - u^2), (assuming all variables are positive) substitute u = a sin(u) and du = a cos(u) du. Then sqrt(a^2 - u^2) = sqrt(a^2 - a^2 sin^2(u)) = a cos(u) and u = sin^(-1)(u/a):
 = a integral a^3 sin^2(u) cos^2(u) du
Factor out constants:
 = a^4 integral sin^2(u) cos^2(u) du
Write cos^2(u) as 1 - sin^2(u):
 = a^4 integral sin^2(u) (1 - sin^2(u)) du
Expanding the integrand sin^2(u) (1 - sin^2(u)) gives sin^2(u) - sin^4(u):
 = a^4 integral(sin^2(u) - sin^4(u)) du
Integrate the sum term by term and factor out constants:
 = -a^4 integral sin^4(u) du + a^4 integral sin^2(u) du
Use the reduction formula, integral sin^m(u) du = -(cos(u) sin^(m - 1)(u))/m + (m - 1)/m integral sin^(-2 + m)(u) du, where m = 4:
 = 1/4 a^4 sin^3(u) cos(u) + a^4/4 integral sin^2(u) du
Write sin^2(u) as 1/2 - 1/2 cos(2 u):
 = 1/4 a^4 sin^3(u) cos(u) + a^4/4 integral(1/2 - 1/2 cos(2 u)) du
Integrate the sum term by term and factor out constants:
 = 1/4 a^4 sin^3(u) cos(u) - a^4/8 integral cos(2 u) du + a^4/8 integral1 du
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du:
 = 1/4 a^4 sin^3(u) cos(u) - a^4/16 integral cos(s) ds + a^4/8 integral1 du
The integral of cos(s) is sin(s):
 = -1/16 a^4 sin(s) + 1/4 a^4 sin^3(u) cos(u) + a^4/8 integral1 du
The integral of 1 is u:
 = -1/16 a^4 sin(s) + (a^4 u)/8 + 1/4 a^4 sin^3(u) cos(u) + constant
Substitute back for s = 2 u:
 = (a^4 u)/8 - 1/16 a^4 sin(2 u) + 1/4 a^4 sin^3(u) cos(u) + constant
Apply the double angle formula sin(2 u) = 2 sin(u) cos(u):
 = (a^4 u)/8 + 1/4 a^4 sin^3(u) cos(u) - 1/8 a^4 sin(u) cos(u) + constant
Express cos(u) in terms of sin(u) using cos^2(u) = 1 - sin^2(u):
 = (a^4 u)/8 - 1/8 a^4 sin(u) sqrt(1 - sin^2(u)) + 1/4 a^4 sin^3(u) cos(u) + constant
Substitute back for u = sin^(-1)(u/a):
 = 1/8 a^4 sin^(-1)(u/a) + 1/4 u^3 sqrt(a^2 - u^2) - 1/8 a^3 u sqrt(1 - u^2/a^2) + constant
Factor the answer a different way:
 = 1/8 (a^4 sin^(-1)(u/a) + 2 u^3 sqrt(a^2 - u^2) - a^3 u sqrt(1 - u^2/a^2)) + constant
Which is equivalent for restricted u and a values to:
Answer: |= 1/8 (u sqrt(a^2 - u^2) (2 u^2 - a^2) + a^4 tan^(-1)(u/sqrt(a^2 - u^2))) + constant

Hey Cookie!....That is what I started with also (and then deleted)......BUT look what happens when n=5    !  Division by ZERO .....Oh No ! Not THAT thread again !!!  lol

Tan x = 240/372 =.64516

Arctan x = x = 32.82 degrees   = .572 R

Sorry, I am not a member so I can’t give points. After using the calculator, I decided to read some of the questions and answers. This is my second visit and that was my first post. I thought it would be deleted quickly, especially posting on your comments with both of you online.  I didn’t expect you to have a sense of humor about it. It is really cool that you do. I reduced the points because the graph is y=1/x instead of xy=c, which is the same thing if c=1. I was just being consistent with the theme of the post. 

CDD!  This reminds me of my firm’s principal troubleshooter. He was always looking for a cure or a vaccine for CDD, although he called it Contagious Dumbness Disease.  He would select anyone who f’d up as candidate for experimental trials.  It was a joke, but he often acted as if it was serious.  

I have a mostly empty bottle of vodka, a half-empty bottle of gin, and a partly empty bottle of bourbon. These are leftovers from the New Year’s Day party.   I have been draining them but it may be a few weeks or so until they are empty. I probably can’t meet the deadline unless I run into more than the average amount of stupid.  LOL I still have to give a good answer too.

65 x 0.8 =52 meters - his distance upstream.

Cos(30) =52 / H

H = 52 / 0.8660254

H = 60.04 meters - the hypotenuse

Width of the river =W^2 =60.04^2 - 52^2

W = 30.02 meters - the width of the river.

Moderators: Please verify this. Thanks.