CPhill

According to WolframAlpha......no solutions exist for this system......

Out of every100 people, 20 now use public transport and 80 use cars.....

And of the 80 that use cars, 30% of these will switch to public transportation....thus.....the public transportation system will increase by 80 * .30  = 24 people out of  every 100

However, it will also lose 10% of its current ridership to cars = 20 X .10   = 2 people out of every 100

So, the percentage of people using public transportation in the long run will be

[20 + 24 - 2] / 100 =  42%

(1/2)x=5/6        multiply both sides by the reciprocal of 1/2   = 2/1

x = (5/6) * 2/1     =  10/6    =  5/3

We want to chose 2 of the 5 red,  2 of the 7 green and 2 of the 6 yellow......so we have :

C(5,2) * C(7,2) * C (6,2)   = 3150 ways

50% of 210   =

.50 x 210   =

105

Here's another way to do this without using too much Calculus

Note that.....at ponit B, a line through (2,0) can be constructed that will be perpendicular to the tangent line on y = sqrt(x) at point B.......and the shortest distance bettween a point and a line is a perpendicular line drawn from the point to the given line

Where sqrt(x)  is differentiable, the slope will be 1 /[2sqrt(x)]

So, the slope of a perpedicular line will be -2sqrt(x)

And the equation of such a such a line through (2,0) will be

y = -2sqrt(x)(x - 2)   ......setting this equal to y = sqrt(x), we have

-2sqrt(x)(x - 2)  = sqrt(x)

-2sqrt(x)(x - 2) - sqrt(x)  = 0

sqrt(x) [ -2(x - 2) - 1]  = 0

sqrt(x) [ -2x + 3]  = 0

So......sqrt(x) = 0   and x =  0 [reject]      or    -2x + 3 = 0   and x = 3/2

This is the x coordinate of B..... and .the y coodinate of B  = sqrt(3/2)  

Actually.....this is an interesting problem......

We can do better than 64 trees if we consider another configuration.....

Since the acre can be in any shape......consider one that is in the shape of an equilateral triangle

To find the side,S, of such a triangle, we can solve this:

43560  = [sqrt(3)S^2/4]  

And S  = about 317.7 ft on each side

Since the trees have to be 25 feet apart, it we plant the first one at the apex, we can actually plant

300/25 + 1   = 12 + 1 =  13 down each side  .....there will be an "dead" space at the bottom of this triangle where no other trees can be planted with a 25 foot spacing requirement......

Then, theoretically, the number of trees that we could plant  = [13*14] / 2   = 91

And, unlike a square planting area, each tree is guaranteed to be 25 ft from any of its "nearest neighbors"........[the diagonal neighbors in a square planting are sqrt(2)*25 ft apart....this "wasted" space is why we can't plant as many as in the equilateral configuration ]

Look at the picture of the first five rows :

Note that each tree is exactly 25 ft from any of its neighbors......!!!!

It would be interesting to see if another arrangement [such as a hexagonal one] would even be more productive.....!!!

sinC / c   = sin B  / b        so we have

sin(114) / 41 = sinB/  28

sinB =  [28* sin (114))/ 41]

sin^-1 [28* sin (114))/ 41]  = B  = about 38.6°

And angle A = about [180 - 114  - 38.6]°   = about 27.4°

And side a  =

a/sin(27.4)  = 41 / sin(114)

a = 41*sin(27.4) / sin(114)   =   about 20.7 units

"Smiles"  is the longest word in the dictionary.......there's a "mile" between each "s"........!!!!!!!

(c) sin(2x) − 4 sin(x) = 0     simplify

2sinxcosx - 4sinx = 0      factor

2sinx [ cosx - 2] = 0        the second factor has no solution

So

2sinx = 0      divide both sides by  2

sinx =  0      and this occurs at  [0 + n*180] °   where n is an integer

(d) 4cos2(θ)−cos(θ)−2=0       simplify

4[2cos^2 (θ) - 1] - cos (θ)  - 2  = 0

8cos^2 (θ) - cos (θ)  - 6   = 0

Let x = cos(θ)

8x^2 - x - 6  = 0

Using the quadratic formula, the solutions to this are  x = [1 - sqrt(193)] /16    and x = [1 + sqrt(193)]/ 16

So

cos (θ)  =  [1 - sqrt(193)] /16       or    cos (θ ) = [1 +sqrt(193)] /16

Taking the cosine inverse to find both angles, we have

(θ)  =   about 143.685°     and (θ)  = about 21.444°  

There are also two more angles on (0, 360)  at about 216.3°   and about  338.6°

Here's the graph of this one......it has an odd periodicity.......https://www.desmos.com/calculator/pwgqsyrurd