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I don't know Chris,

This might be the answer wanted but

Newton's law of cooling says that 

$$\\\mbox{T is temperature, t is time, and S is the surrounding temperature}\\\\
\frac{dT}{dt}=-k(t-S)\\\\
$I think this means$\\\\
T(t)-S=(T_0-S)e^{-kt}\\\\$$

I don't think that we are given enough information to solve this question.   

$${\mathtt{a}} = {\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$     

 find      $${{\mathtt{a}}}^{{\mathtt{2}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = \left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left(\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right)\right)}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left({\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{11}}\right)}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{25}}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{2}}$$

and that means that   $$a=\pm \sqrt{ 2}$$          COOL !!!

$${\frac{{\sqrt[{{\mathtt{{\mathtt{7}}}}}]{{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{{\mathtt{a}}}^{{\mathtt{4}}}}}}}}{{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{\mathtt{a}}}}}}$$

$$\\=\frac{\sqrt[7]{a^2\times a^{4*(1/5)}}}{a^{1/5}}\\\\
=\frac{\sqrt[7]{a^2\times a^{0.8}}}{a^{0.2}}\\\\
=\frac{\sqrt[7]{a^{2.8}}}{a^{0.2}}\\\\
=\frac{(a^{2.8})^{1/7}}{a^{0.2}}\\\\
=\frac{a^{2.8/7}}{a^{0.2}}\\\\
=\frac{a^{0.4}}{a^{0.2}}\\\\
=a^{0.4-0.2}\\\\
=a^{0.2}\\\\
=a^{1/5}\\\\
=\sqrt[5]{a}$$

That is ok.   We are here to teach so if you still have problems we encourage you to tell us.

I have now included the final answer.  If you still have concerns let us know :)

I have now included the final answer as you requested.

If you still have a problem let me know and I will try to talk you through it :)

That is really cool.  Thanks Chris :)

I am going to add this thread into  the      Sticky Topic :  Reference Material     :)

thanks Chris.  

the slope is the gradient :)

Thanks Chris :)

Yep....nice job !!!

It can be shown that the curve traced out by a point on this wheel  is known as a cycloid. It's arc length through one rotation can be found by using Calculus and is equal to 8r, where "r" is the radius of the wheel....!!! Also, the area under one arc is just .... 3*pi*r^2......

y+3<-2/3(x-9)  subtract 3 from both sides

y < -2/3(x-9) - 3   simplify

y < (-2/3)x + 6 - 3

y < (-2/3)x + 3

A= 0.15 (d - 7,550) + 755

So

3455 = .15(d - 7550) + 755   subtract 755 from both sides

2700 = .15(d - 7550)     divide both sides by .15

2700 / .15   = d - 7550    add 7550 to both sides

2700 / .15   + 7550  = d = $25550

3 - (k + 1) = -2      add (k + 1) to both sides

3 = -2 + (k + 1)     add 2 to both sides

5  = k + 1              subtract 1 from both sides

4  = k

Thanks anon,

That is an approximation of pi.  

If you roll a wheel one complete rotation then it will travel exactly  $${\mathtt{\pi}}$$  times the diameter of the circle in distance.

pi is an irrational number which means that it cannot be expessed exactly in digits.

It is approximately  3.14158 

[I give credit to flflvm97 for first using this Gif on the forum.  Thanks fl :) ]

Wow, now I understand! thanks. 

Whatever n+1 and n-1 are, they are two integers apart....

So......if we divided (n+1)!  by (n-1)!  the result  = (n+1)* n

So we have

(n+1)*n = 90

n^2 + n = 90    subtract 90 from both sides

n^2 + n - 90  = 0  factor

(n -9) (n +10)  = 0      so, either n - 9 = 0   or n + 10 = 0

And, since were assuming a positive integer, n = 9

So

(9 +1) ! / (9 - 1) !  =

10! / 8!  =

10 * 9  =

90

We can set this up as two vectors, the first for the plane and the second for the wind...so we have

u = < 500cos310, 500sin310 >      v = < 45cos225, 45sin225 >

Adding the x components, we have 500cos310 + 45cos225 = 289.57

Adding the y components, we have 500sin310 + 45sin225 = -414.84

The resultant is given by √(289.57² + (-414.84)² ) = about 505.9 km'hr

The direction is given by tan^-1(-414.84/289.57) = about (- 55.08) degrees  = about 304.91 degrees

This makes sense......the wind is helping to increase the speed of the plane but it is also blowing it toward a more "southerly" direction.....

OK.....

Thanks Chris :)

This cannot be simplified unless you have values for a or b or c :)

Everyone has their own abilities......yours are probably different from another's........constantly comparing yourself to others is a losing game.....play to your strengths, work on your weknesses.....that's all I've got......

There are infinite possibilites for this....

The equation will be in the form

x^2 / 121  - y^2 / b^2   = 1        where "a" = 11 and a^2  = 121

I'm going to use the following function to model this

T = ab^x      where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab²     →   323/b²  = a

And when x = 5   we have

288 = ab⁵  =  (323/b²)b⁵   = 323 b³     divide both sides by 323

288/323 = b³    take the 3rd root of each side

(288/323)^(1/3)  = b = about .9624

So, a = 323/(.9624)² =  about 348.66

So, our function is

T = 348.66(.9624)^x

So, to find out when they will be at 120 degrees, we have

120 = 348.66(.9624)^x        divide both sides by 348.66

120/348.66  =  (.9624)^x    take the log of each side

log ( 120/348.66 )  = log (.9624)^x     and by a log property, we have

log ( 120/348.66 )  =  x log (.9624)    divide both sides by log(.9624)

log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes

I am just thinking here.

If we increase something by a factor of 1 this is meaningless.  To a mathematician it means multiply by 1 so it stays the same and is not increased at all.

If you told a lay person in the street to increase the size of something by a factor of 1 they could either mistakenly add one unit to it or erroneously double the size. They could easily think that if it was 4 units long before and increase it by a factor of 1 that it must now be 8 units (what it was before plus one more of what it was before 1+1more) .  Of course this is really increasing by a factor of 2 (1*2) but I can see how it could be very confusing.