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what is cos(15)divided by 6

\(\large \frac{cos(15deg)}{6}=\frac{0.9659}{6}=0.16099\)

\(\large \frac{cos(15rad)}{6}=\frac{-0.7597}{6}=-0.12661\)

  !

Guest, would you create an account and message me, if you please?  I may very well be acquainted with your “Troubleshooter”

0.1609876377

Math Jokes; LOL

How many more tricycles do the shipments contain than four wheel (training) bicycles?

Let b = number of 2 wheel bikes, t = 3 wheel trikes, f = 4 wheel bikes

b + t + f = 95

2b + 3t + 4f = 250

b = t + 20

You now have three independent equations in three unknowns.  Can you take it from here?

EP, why did you use a slide rule?  Scientific Calculators were very common and cheap then, and a few of the programmable calculators cost about the same as an expensive text book. 

Did you go to an Amish school?

Comments from SevenUp

X is the result

"0b" idicates binary. Hex is "0x".

Time to test my genetic enhancement.

\(\text {Velocity of hammer at impact with pile}\\ V_f= \sqrt{2ad} \hspace{1em}|\sqrt{2*9.81 m/s^2*5m} = 9.904568m/s\\ \)

\(m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\leftarrow \text {conservation of momentum}\\ 150kg (9.905m/s) = \underbrace{(150 + 400)kg}_{\text{mass of pile & driver}} (v)\leftarrow \text {velocity of combined masses}\\ \hspace {7em} v=\dfrac{150kg (9.905 m/s)}{(150 + 400)kg} = 2.70136m/s\\\)

\(\text {Energies:}\\ \text{kinetic energy} = \frac{1}{2}(550 kg)(0 - 2.70136 m/s)^2 = - 2006.77 J\\ \text{potential energy} = (550 kg) (9.81 m/s^2)(0-0.075 m ) = -404.66 J \\\)

\(\text {Energy to work formula:}\\ \text{(-F)}*(0.075 m) = -2411.43 J \hspace{1em} | (-F) = \text{negative Force} \\ \text{(-F)}=\dfrac{-2411.43J}{0.075m}= 32152.40 \ newtons\\\)

\(\text{Average resistance of the ground }= 32,152.40 \ newtons\)

No figs. Just newtons on this ground. 

This concludes the test of my genetic enhancement.

I think I passed.

Also.....

36  + 5/8   =

36.625

I think, as the question is written......you're answers are correct.....AND you are correct again when you say the Q is confusing !   

What is 36 5/8% to 100?

\(100\times 36\frac{5}{8}\%=\frac{100\times 36\frac{5}{8}}{100}=36\frac{5}{8}\)

!

Whatever....this question is confusing......!!!!

I see now.....they start at the high point when theta = 0  originally     so they will be at 25 above center and never reach 15cm above that!      d'Oh!

NOW I am more confused     h = 25 cos(5theta)  is it's position from its CENTER......

 Is that what you assumed , Chris??

Disregard all of this......time to rethink my answers !!!!!

A) At what rotation angles will the horse be displaced 15 cm above its original position in one revolution? 

The horse is at its max height [25cm] when the ride starts  [at θ  = 0] ......thus, it never rises15 cm above its original poisition.......

B) At what rotation angles will the horse be displaced 20 cm below its original position in one revolution?

The height at those angles will be 5cm....so we have

5  = 25cos(5θ)  

Have a look at the graph here  [in degrees].......https://www.desmos.com/calculator/pusq0nejlq

The graph shows that this height occurs 10 times in one revolution

C) How long will it take for one complete revolution if the carousel rotates at a speed of 24degrees/s?

360 / 24  = 15 seconds

A   5 times every rotation is every 72 degrees

   10.62 + 72 =   82.62        154.62     226.62    298.62   370.62

   BUT, they also will be displaced 15 cm on the way DOWN for another 5 times each rotation

     arcos(-15/25) = 5t = .44 R    25.37 degrees  97.37   169.37  2.41.37  313.37  degrees

(I think !)

Ooops sorry....I gave onlt the FIRST position of those displacements.....back in a few

A:   15cm = 25cos(5t)      t= theta

     15/25 = cos 5t

     Arcos (15/25) = 5t      t= .185 R =10.62 degrees

B  -20 = 25 cos(5t)

    -20/25 = cos 5t

  Arcos (-20/25) = 5t    t = .4996 R   28.62 degrees

C   360 degrees/ (24 degrees/sec)  = 15 seconds

What grade is he in lol

Ah...I see what you mean, EP....I was taking it as the cumulative distance........4.9 ft in one second .....14.9 ft  total after two seconds, etc......I think your interpretation is correct.....