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Ok, 60 do both drama and sports, so 85-60=25 only take drama, and 200-60=140 on do sports. Adding, 140, 25, and 40, we get 205- do everything. But, the question is asking, what is the probability, that a student chosen at random is in neither.  This is just, 330-205=125. So, 125/330*100=37.87

The answer is \(\boxed{41}\)

1)

1- 3cosA - 4cos^2A                 1  - 4cosA

________________      =     ___________

       sin^2A                               1  - cos A

Note that we can factor the numerator on the left as

(1 - 4cos A) ( 1 + cos A)

___________________

  1  - cos^2A

(1 - 4cos A) (1 + cos A )

____________________

(1 - cos A) ( 1 + cos A)

1 - 4cosA

_________       which equals the right side

1  -  cos A

2)

tan A   =    1 + tan A

                 ________

                 1 +  cot A

One trick to try here, AT, is to write the right side in terms of sines and cosines

So we have

1 +  sin A /  cos A

_______________       get common denominators on top/bottom

1  + cos A / sin A

[cos A + sin A} / cos A

___________________      invert the bottom  fraction and multiply

[ sin A + cos A]  /  sin A

[ sin A + cosA ]           sin A

_____________   *    ____________

     cos A                   [  sin A + cos A]

sin A       *    [ sin A + cos A]

____             ____________

cos A          [ sin A + cos A ]

sin A       *  1

____

cos A

tan A  *  1   =   tan A     which is the left side     

Let  27 cm  be the base of triangle CDE .

Let  h cm  be the height.

sin( angle )   =   opposite / hypotenuse

sin 32°  =   h / 14                                         Multiply both sides by  14 .

14 sin 32°   =   h

area of triangle CDE   =   (1/2)(base)(height)

area of triangle CDE   =   (1/2)(27)(14 sin 32°)

area of triangle CDE   =   189 sin 32°

area of triangle CDE   ≈   100.155

By the law of cosines...

EC²   =   14² + 27² - 2(14)(27) cos 32°

EC²   =   196 + 729 - 756 cos 32°

EC²   =   925 - 756 cos 32°

area of sector / area of circle  =  105° / 360°

area of sector   =   105° / 360° * area of circle

area of sector   =   105° / 360° * pi * EC²

area of sector   =   105° / 360° * pi * (925 - 756 cos 32°)

area of sector   ≈   260.115

area of logo   =   area of sector + 2(area of triangle CDE)

area of logo   ≈   260.115 + 2(100.155)

area of logo   ≈   460.425

And  460.425  rounded to the nearest ten   =   460   

Wow, your amazing CPhill! Thank you!

Note...tertre....if we find the center of the circle and draw a radius  parallel to the bottom base of the large equilateral triangle, we will create another smaller equilateral triangle with a side of 6 cm

So.....the area of the sector formed by the two radial sides of this smaller equilateral triangle  will equal 1/6 of the circle whose radius  = 6  =

(1/6)pi (6^2)  =  6pi  cm^2        (1)

And the area of this smaller  equilateral triangle = 

(1/2)(6^2)*sqrt (3) / 2  =  

36sqrt (3) /4  =

9sqrt (3)  cm ^2       (2)

So....the   area of one of the shaded regions  is   (1)  - (2)  =

{ 6pi - 9sqrt (3)]  cm^2

So...by symmetry.....both shaded regions have an area of

2 [ 6pi - 9sqrt (3)]  cm^2  =

[ 12pi  - 18sqrt (3) ]    cm^2     

So

a + b + c  =

12 + 18 + 3

33

1) 

2sin^2A - 1  =   sin^4A - cos^4A

Working with the right side we can factor this as

(sin2^A + cos^2A) (sin^2A - cos^2A)  =

(1) ( sin^2 A - cos^2A)  =

sin^2A  -  (1  - sin^2A) =

sin^2A  - 1 + sin^2 A  =

2sin^2A  - 1        which is = to the left side

2) 

[ cos^2A - csc^2A + sin^2A]

     _______________________   =  - cot^3A

          sinAsec A

Obviously...the left side is the side that needs to be simplfied

Note......  sin^2A +cos^A  = 1

  [  1  - csc^2]

______________

   sinA (1/cos A]

Note

cot^2A  + 1  = csc^2A   ....so...

1 - csc^2A = - cot^2A

-cot^2A  * cosA

____________

   sinA

-cot^2A  *  cosA

                 ______

                  sin A

-cot^2A * cot A

-cot^3A       which is the right side

There are  46 total  students  in the class

Those that play football only  =  14 - 8  =  6

Those that play cricket only  = 20 - 8  =  12

So....those that play football,  cricket or both  = 6 + 12 + 8   =  26

So...the porbability that a student plays football, cricket or both is

26 / 46  =

13 / 23

We can find angle BCA  in the large triangle thusly :

sin BCA / 5.6  =  sin 50  / 9.3

sin BCA  =  5.6sin 50  / 9.3

So

arcsin (5.6 sin 50 / 9.3)  = BCA  =  27.47°

So....in the smaller triangle BDC, angle BDC  is supplemental to angle BDA  =  180 - 78  = 102°

So  x  = angle DBC  =

180 - BDC - BCA  =

180  - 102 - 27.47  = 

50.53° =  50.5°

13 degrees  17 minutes 50 seconds

We don't need to worry about the degree conversion..we need to convert the minutes and seconds to decimal degrees

There are 60 minutes to each degree  so we can express this as     17/60  degrees

And there  are  3600 seconds in each degree so we can express this as  50/3600 degrees 

So...adding   D, M ,S  in terms of degrees, we have...

D   +  M     +   S     =

13°  + 17/60° + 50/3600°   =

13.297222°  =   13.297°  (rounded)

BTW, Manuel....here's an app where you can check these conversions :

https://www.rapidtables.com/convert/number/degrees-minutes-seconds-to-degrees.html

However....if you use this fractional adding procedure above.....you shouldn't have much trouble on these!!!

\(\begin{align*} \frac{1}{10}[(1\cdot 2)&+(1\cdot 3)+(1\cdot 4)+(1\cdot 5)+{}\\ &(2\cdot 3)+(2\cdot 4)+(2\cdot 5)+(3\cdot 4)+(3\cdot 5)+(4\cdot 5)]\\ &\qquad\qquad\qquad\qquad=\frac{85}{10} = \boxed{8.5}. \end{align*}\)

This object  is known  as the frustum  of a right circular cone .

The volume is given by :

pi * height  ( r₂^2 + r₁r₂ + r₁^2 ) / 3

The height is not given but it is the leg of a right triangle  with the other leg  = 12 - 8  = 4

And  the hypotenuse = l  = 10

So  the height  is sqrt (10^2  - 4^2)  =  sqrt (100 - 16)  =  sqrt (84)  = 2sqrt (21)

So the volume is

pi * 2sqrt (21) * [ 12^2  + 8*12 + 8^2 ]  / 3  =

pi * 2sqrt (21) * [ 144 + 96 + 64] / 3  =

pi * 2sqrt (21) *  [ 304] / 3   =

[ (608/3) * sqrt (21) pi  ]    units ^3

thanks!

Here's a resource that also might help. Manuel :

https://www.mathsisfun.com/sine-cosine-tangent.html

Angle ADB  = 90°

And angle DBA = 71°

So  angle BAH = 90 - 71  =  19°

And angle BEA  = 90°

And angle EAB  = 46°

So  angle ABH = 90 - 46  =  44°

So...in triangle AHB,  angle AHB  =  180  - 19 - 44  = 117°

Here's a pic  :

67 / 2010 = 1/30. But then it is a repeating decimal!!

201 / 2010 =0.1 This is a terminating decimal.

No problem, 

Thank You all

I believe it's -1:1

I believe it is adjacent side to the hypotenuse. (soh cah toa) 

\( 4(x^4 + 3x^2 + 1)\)  =

4x^$ + 12x^2  + 4

4^2*(2)  + (12)^2 

16*2  + 144

32 + 144

176 

Since we know triangle ABC is a 15-20-25 right-angle triangle, then we can easily  find the x coordinate of the second intersection of AC  and this circle as follows:
(150*2) / 25^2 =12/25 x 15 =7.2
(150*2) / 25^2 =12/25 x 20 =9.6, then D =7.2 and 9.6 and:
BD^2 = 7.2^2 + 9.6^2 
BD^2 =144
BD =12

I don't know of any name, but here's a derivation:

Ah thank you for the response! I didn't know whether it wanted the expression in simplest form.....but since it doesn't, your answer is correct! 

So...

We are trying to find the maximum value of the function \(Y(p)={900p\over2+p}\) within the domain \(x\ge0\). This is the graph of a hyperbola, and it is positive. So, in order to find the maximum value of the function, we have to get its horizontal asymptote. An asymptote is a line that a graph approaches but never touches.  The rules are as follows:

1. If the numerator's degree is less than the denominator's degree, then the horizontal asymptote is y = 0.

2. If the numerator's degree is equal to the denominator's degree, the the horizontal asymptote is the quotient of the coefficients of the highest degree terms.

3. If the numerator's degree is greater than the denominator's degree, then it does not have a horizontal asymptote.

The degree of 900p is 1, and the degree of 2+p is 1. Therefore, the horizontal asymptote is \(y={900\over1}\), or just \(y=900\). This means that the maximum value of the function is 900.

What this actually means is that the function approaches 900 as you go farther along the x-axis, or, as you increase the amount of pesticide. So, theoretically, if you had infinite pesticide, you would have 900kg of extra yield. But... infinity is tricky. Just trust me, the answer is 900.