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Picking up from guest...

at 5 minutes     y = 40 (5) = 200 meters        x = 30(5) = 150 meters       h = 250   (via pythagorean)

x^2 + y^2 = h^2   

     2x dx/dt       +   2y dy/dt   = 2h dh/dt

2(150) (30)  +   2(200)(40)   = 2(250) dh/dt

dh/dt = 50 m/sec                                                   (NOT positive !)

Call  x  the number of weeks it will take  for  both cards to  have the same number of  punches.....we have

5 + 3x   =  4  +  4x      

5 -4  =  4x - 3x

1  =  x   

It will take one week  and  each card will have   5 + 3(1)  =   8 punches

                                                    B

                        A      15               X      5      C

If  BXA   = 90   then so   is  BXC

Note  that triangle  ABC ≈ AXB ≈ BXC

And  we  have this realationship

AX / BX   =   BX  / CX

So

15/ BX =   BX / 5     cross-multiply

15 * 5  = BX^2

75  = BX^2       take both roots

sqrt ( 75)  = BX

sqrt (25) * sqrt (3)  = BX

5sqrt (3)  =  BX

A P P L E = 5! / 2==60 PERMUTATIONS:

APPLE, APPEL, APLPE, APLEP, APEPL, APELP, ALPPE, ALPEP, ALEPP, AEPPL, AEPLP, AELPP, PAPLE, PAPEL, PALPE, PALEP, PAEPL, PAELP, PPALE, PPAEL, PPLAE, PPLEA, PPEAL, PPELA, PLAPE, PLAEP, PLPAE, PLPEA, PLEAP, PLEPA, PEAPL, PEALP, PEPAL, PEPLA, PELAP, PELPA, LAPPE, LAPEP, LAEPP, LPAPE, LPAEP, LPPAE, LPPEA, LPEAP, LPEPA, LEAPP, LEPAP, LEPPA, EAPPL, EAPLP, EALPP, EPAPL, EPALP, EPPAL, EPPLA, EPLAP, EPLPA, ELAPP, ELPAP, ELPPA, >Total distinct permutations = 60

thx :)

172 cents / 9 envelopes  = ~ 19 cents per envelope

pretty sure this quuestion is asking for a rate and not a set distance. "how fast is the straight distance between then changing five minutes later?"

xy - z = 2

xz - y = 5

yz - x = 10

What xyz?

Use substitutions to get:  x=2,  y =3,  z =4

[xyz] =[2  x  3  x  4] = 24

Pythagorean theorem

(40(5))²   +  (30(5))²  = distance ²     (NOTE : distance ² )

27*a mod 40 =17, solve for a

a == 40n + 11, where n=0, 1, 2, 3........etc.

The smallest "a" =11 and the 2nd smallest "a" =40+11 =51

Thanks for trying but the answer is 6!

[5 C 3 * 5^2] + [5 C 4 * 5^1] + [5 C 5 * 5^0]=[250 + 25 + 1] =276 * 6 =1656 / 6^5 == 23 / 108

There are only two integers that work, x = 1 and x = 2.

Thanks for trying my problems

Hi Old Timer it is good to see you again.

Let R be the outer radius at any given point in time

Ler r be the inner radius at any given point in time

Let k be the length which is a constant.

Volume is also a constant so we can work that out first.

\(Volume=(\pi R^2-\pi r^2)k\\ Volume=\pi k (R^2-r^2)\\ When\;\; R=0.05\;\; r=0.03\\ so\\ Volume=\pi k (0.0025-0.0009)\\ Volume=0.0016k\pi\\\)

Now I will need to know what 

NOW:

WE need to find   

\(\frac{dR}{dt}\quad when \quad \frac{dr}{dt}=0.5mm/sec\)

\(\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\)

So I need to find dR/dr

\(V=\pi k (R^2-r^2)\\ 0.0016k \pi =\pi k (R^2-r^2)\\ 0.0016= (R^2-r^2)\\ R^2=0.0016+r^2\\ R=(0.0016+r^2)^{0.5}\\ \frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\ \frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\ \frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\\)

\(\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\ When\qquad \frac{dr}{dt}=0.5\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\ \)

r = 0.03   we were told that in the question.

\(\frac{dR}{dt}=\frac{0.03}{\sqrt{0.0016+0.0009}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{\sqrt{0.0025}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{0.05}\times 0.05\\~\\ \frac{dR}{dt}=0.03 mm/sec\\~\\\)

If I have not done anything careless then I think my answer is ok but maybe I turned a molehill into a mountain.... 

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LaTex:

Volume=(\pi R^2-\pi r^2)k\\
Volume=\pi k (R^2-r^2)\\
When\;\; R=0.05\;\; r=0.03\\
so\\
Volume=\pi k (0.0025-0.0009)\\
Volume=0.0016k\pi\\

V=\pi k (R^2-r^2)\\
0.0016k \pi =\pi k (R^2-r^2)\\
0.0016= (R^2-r^2)\\
R^2=0.0016+r^2\\
R=(0.0016+r^2)^{0.5}\\
\frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\
\frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\
\frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\

\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\
\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\
When\qquad  \frac{dr}{dt}=0.5\\

\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\

\(2\cos\theta+\sin(2\theta)=0\)

Let's use the double-angle formula for sin which says  \(\sin(2\theta)=2\sin\theta\cos\theta\)

\(2\cos\theta+2\sin\theta\cos\theta=0\)

Now we can factor  2 cos θ  out of both terms on the left side of the equation.

\(2\cos\theta(1+\sin\theta)=0\)      (Notice if we distribute  2 cos θ  we get the previous expression)

Set each factor equal to  0  and solve for  θ:

\(\begin{array}{ccc} 2\cos\theta=0&\text{or}&1+\sin\theta=0\\~\\ \cos\theta=0&\text{or}&\sin\theta=-1\\~\\ \theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\dots&\text{or}&\theta=\frac{3\pi}{2},\frac{7\pi}{2},\frac{11\pi}{2},\frac{15\pi}{2},\dots\\~\\ \theta=90^\circ,270^\circ,450^\circ,630^\circ,\dots&\text{or}&\theta=270^\circ,630^\circ,990^\circ,1350^\circ\dots \end{array}\)

The solutions in the interval  [0°, 360°)  are:   90°, 270°

I'm using  x instead of  theta   (same difference)

2cosx  - sin 2x   =   0

2cos x - 2sinx cosx    =  0     factor

2cos x ( 1  - sinx )  =  0        set each factor to 0  and solve

2cos x =  0    divide through  by  2                           1  - sin x  =  0

   cos x  =  0                                                               sin x =  1

x =  90°, 270°                                                              x  = 90°

So   x =  90° , 270°

5 - 2 - 5    =    3 ways

5 - 1 - 6    =    6 ways

4 - 4 - 4     =   1  way

3 - 5 - 4    =    6 ways

3 - 6 - 3   =     3 ways

2 - 6  - 4  =     6 ways

There are  216  possible outcomes

P ( 12)  =  25/ 216

This never will happen

The time to  max  height  =  -60/ (2 *-16)  =  60/32 =  1.875 sec

And  the height  will be   

-16(1.875)^2  + 60 (1.875)    = 56.25  ft

x is   (-1,1) 

numerator    x-1   is between    -2 and 0

denominator x-3   is between    -4  and -2

-2/-2 = 1

0/anything is 0

So  the domain of L  is  (0,1)

             x^4     + x^3    +   x^2    + x  + 1

1-x  [   -x^5   + 0x^4  +  0x^3  + 0x^2   +  0x  +  1 ]

            - x^5   + x^4

           _____________________________________

                        -  x^4   +  0x^3

                        -  x^4   + x^3

                       ___________________

                                   -  x^3   +  0x^2 

                                   -  x^3    +  x^2

                                    _________________

                                                -  x^2   +  0x

                                                -  x^2    + x

                                                ___________

                                                             -  x   +  1

                                                              - x   +  1

                                                          ___________

Notice   that   ( x^4  + x^3 + x^2  + x) + 1   =  5 +  1   =   6

The Extreme Value Theorem says  that  on  a bounded interval,  a  continuous function  must reach  a maximum and a minimum  at least once

y = x    is  not   bounded  function   because it has no max or  min over its domain   [ we could  restrict the  domain to make it bounded  ]

A=  (3,3) =  (x1, y 1)

D = (2,9)  = (x2, y2)

C = (-4,3) = (x3, y3)

B = (-3,-3)=  (x4, y4)

Area = (1/2)  [ ( x1y2  + x2y3 + x3y4  + x4y1)  - (x2y1 + x3y2 + x4y3 + x1y4 ) ]  =

(1/2)  [  (27 + 6  + 12  -9)  - ( 6 -36 - 9  -9)  ]  =

(1/2)  [ 36  + 48  ]  =   84/ 2  =   42

Area of triangle EFB  =(1/2) (3.5) (3)  =  5.25 

P ( below  the  x axis )  =   5.25  / 42  =  ( 21/4)  / 42  =  1/8

1.645

First let us find the equation for the lines of AB and CD (the lines that cross the x axis). Then, we can find the area of the region under the x axis and put that over the area of total area. Probability is success/total, and in this case, the area under the x axis is "success", and the total area is the "total."

Line AB has points (3,3) and (-3,-3). Writing this in slope-intercept form, which is y=mx+b, we get:

3=3m+b

-3=-3m+b

Adding these equations we get:

0=2b, b=0.

Substituting b=0 into our original equations we get:

3=3m+0, m=1.

Therefore, the line for AB is:

y=x.

Now it is seemingly obvious that when the line crosses the x axis (when y is 0), it passes through point (0,0).

We can repeat the same steps we did for the first line for the second, and we will find that the equation of the second line (CD) is:

y=2x+5

Therefore, this line crosses the x axis when y is 0 and x is -5/2.

From this information, we can now find the area of the region under the x axis. This is a funky shape, so it would be easiest to draw a picture, find the area of a big rectangle around the figure, and subtract triangles until you get what is left. I found that the area of the region was 6.75, and the are of the big figure was 28.5.

Therefore the probability is 6.75/28.5 = 9/38.  <-- My Answer