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Need to see a pic, bambam.....  

No prob !!!

Thank you very much kind soul! 

In 3 seconds, the distance that the unicycle travels =

2 revs * distance traveled per rev  =

2   * 2pi (9)  =  36 pi   inches

So...in 1 second  it travels  1/3 of this =    36 pi / 3    =   12 pi inches / sec ≈ 37.7 inches / sec

Supercalifragilisticexpialidocious!!!!

hi mirb u cheat on number theory aops hw, u ————

Thanks, EP.....

It might look like this :

Good, Synth....and you  ???

1.    Y = 2 X  + 20

2.    Y = 3 X   

Here's the graph:

Pythagorean theorom again

a^2 + b^2 = c^2

Hypotenuse is the diagonal measure of 113

One leg is 112

 a^2 + 112^2 = 113^2

a^2 = 113^2 - 112^2

a = 15 in

thank you!!

Use the pythagorean theorem

   hypotenuse is the ladder   58 ft

   one leg is the building side  42 ft

    find the other leg

       a^2 + b^2 = c^2

     a^2 + 42^2 = 58^2

      a^2 = 58^2-42^2

       a = 40 ft

Thank you!

5x+8y= -9

8y = -5x-9

y= -5/8 x - 9/5     slope = m = -5/8

Perpindicular slope =  8/5

L2   y = 8/5  x  + b

       10 = 8/5 (10) + b

           b = -6

L2  y = 8/5 x -6          m+ b =  8/5  - 6 = -22/5

Do you have the referenced diagram? 

Men to Children =  11/2

Women to children 10/3

men to women  =  11/2  /  10/3 =  11/2  *  3/10 = 33/20

Total fish in lake = 10 + 23 + 39 = 72

Percentage perch   10/72 = 13.88%   13.88% of a complete circle pie chart

   is  .1388 x 360 degrees = 50 degrees

Bream  23/72 = 31.94%      = 115 degrees

Carp   39/72                          = 195 degrees  on the pie chart

TOTAL should equal 360 degrees

    as a checkkk 50 + 115 + 195 = 360    Check!

        Now just draw out the pie chart with the sectors as above    ......

            It appears you have drawn in the 195 sector for carp.....now draw in the 50 and 115 sectors in the remaining space on the pie chart for perch and bream respectively.....

Men to Children = 11:2  multiply both by 3 =33:6

Women to Children =10:3 multiply both by 2 =20:6

Men to Women =33:20

As follows:

Thank youuuuu

-Given that 2a = 3b show that tan theta  = 4/7 and find the corresponding values of theta

\(\begin{array}{|lrcll|} \hline &&& \boxed{a= 2\sin(\theta) + \cos(\theta) \\ b= 2\cos(\theta) - \sin(\theta) } \\\\ &2a &=& 3b \\\\ &2\Big(2\sin(\theta) + \cos(\theta)\Big) &=& 3\Big( 2\cos(\theta) - \sin(\theta)\Big) \\\\ &4\sin(\theta) + 2\cos(\theta) &=& 6\cos(\theta) - 3\sin(\theta) \\\\ &4\sin(\theta) + 3\sin(\theta) &=& 6\cos(\theta) - 2\cos(\theta) \\\\ &7\sin(\theta) &=& 4\cos(\theta) \quad | \quad : \cos(\theta) \\\\ &7\cdot \dfrac{\sin(\theta)}{\cos(\theta)} &=& 4 \quad | \quad : 7 \\\\ &\dfrac{\sin(\theta)}{\cos(\theta)} &=& \dfrac{4}{7} \quad | \quad \dfrac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \\\\ &\tan(\theta) &=& \dfrac{4}{7} \\\\ & \theta &=& \arctan\left(\dfrac{4}{7}\right) + z\cdot 180^{\circ}, ~ z \in \mathbb{Z} \\\\ & \theta &=& 29.7448812969^{\circ} + z\cdot 180^{\circ} \\\\ z=0: & \mathbf{\theta} &\mathbf{=}& \mathbf{29.7448812969^{\circ}} \\\\ z=1: & \theta &=& 29.7448812969^{\circ}+ 180^{\circ} \\\\ & \mathbf{\theta} &\mathbf{=}& \mathbf{209.744881297^{\circ}} \\ \hline \end{array}\)

The corresponding values of theta are:  \(\mathbf{29.7448812969^{\circ}}\) and \(\mathbf{209.744881297^{\circ}}\)

It is given that a=2sin theta + cos theta b=2cos theta-sin theta,

where theta is greater than or equal to 0 degrees and smaller than or equal to 360 degrees.

-Show that a squared +b squared is constant for all values if theta

\(\begin{array}{|rcll|} \hline && \boxed{a= 2\sin(\theta) + \cos(\theta) \\ b= 2\cos(\theta) - \sin(\theta) } \\\\ \mathbf{a^2+b^2} &=& \Big(2\sin(\theta) + \cos(\theta)\Big)^2 + \Big(2\cos(\theta) - \sin(\theta)\Big)^2 \\\\ &=& 4\sin^2(\theta) + 4\sin(\theta)\cos(\theta) + \cos^2(\theta) \\ &+& 4\cos^2(\theta) - 4\sin(\theta)\cos(\theta) + \sin^2(\theta) \\\\ &=& 4\sin^2(\theta) + \cos^2(\theta)+ 4\cos^2(\theta) + \sin^2(\theta) \\\\ &=& 4\sin^2(\theta)+ 4\cos^2(\theta) + \cos^2(\theta) + \sin^2(\theta) \\\\ &=& 4\Big(\sin^2(\theta)+ \cos^2(\theta)\Big) + \Big(\sin^2(\theta) + \cos^2(\theta)\Big) \\\\ &=& \Big(\sin^2(\theta)+ \cos^2(\theta)\Big)(4+1) \\\\ &=& 5\Big(\sin^2(\theta)+ \cos^2(\theta)\Big) \quad | \quad \sin^2(\theta)+ \cos^2(\theta) = 1 \\\\ &=& 5 \cdot 1 \\\\ &\mathbf{=}& \mathbf{5} \\ \hline \end{array} \)

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