CPhill

This is impossible.....two sum of any sides of any triangle is always greater than the remaining side.

But  99 + 1212  <  1515

The number of total possibilities is C(10,4 ) = 210

But the sum of any four of the first ten prime numbers is odd only if 2 is included.

So, this means that, out of the other nine, we want to choose any three of them.

So C(9,3)  = 84

So...the probability that the sum of any four of the first ten prime numbers is odd is given by ;

84 / 210 =  28/70 = 14/35

Just divide 43.25 by 12.5......

f(x ) = x^3(x+2)(x-4)

We could use the product/power rules, but I find just expanding this and using the power rule by itself is easier.

So we have

f(x) = x^3(x^2 - 2x - 8)  =  x^5 - 2x^4 - 8x^3

And 

f ' (x) =  5x^4 - 8x^3 - 24x^2

The volume of the cylinder is

V = pi * r^2 * h   where r is the radius of the cylinder and h is the height.

And the radius of the ball = the radius of the cylinder. 

But since the ball touches the sides, top and bottom, the height  of the cylinder must be 2r.

So the volume of the cylinder is

V = pi * r^2 * 2r  =  2pi *r^3

And the volume of the ball is

V = (4/3)pi* r^3

So..... the portion of the cylinder that is filled  by the ball (in terms of a ratio) is

[(4/3)pi * r^3] / [ 2 pi * r^3]  =  (4/3)/2 = 4/6 = 2/3 of the cylinder

So....the empty portion, in terms of ratio  = 1/3 of the cylinder 

23.  Using the Law of Cosines, we have

AP^2  = 15^2 + 12^2  - 2(15)(12)cos 24

AP = about 6.33

24. The diagonals of a rhombus bisect each other, therefore AP = PC  = 6.33

25. Since the diagonals bisect each other, BP = PD = 12. So, BD =24. And we can use the Law of Cosines to find AD

AD^2  = 24^2 + 15^2 - 2(24)(15)cos 24

AD  = about 11.97

So, using the Law of Sines

sin BDA / AB = sin 24 / AD

sinBDA / 15 = sin 24 / 11.97

sin^-1(15 sin 24 / 11.97) = BDA = 30.64°

26. We can use the Law of Cosines to find ACB....note BC = AD = 11.97  and AC = 2(AP) =2(6.33) = 12.66

AB^2 = BC^2 + AC^2 - 2 - 2(BC)(AC)cosACB

15^2 = 11.97^2 + 12.66^2 - 2(11.97)(12.66)cosACB

cos^-1 = (15^2 - 11.97^2 - 12.66^2) / (-2(11.97)(12.66)) = ACB = 74.98°

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8))=

(-3/8) [log₂ 3 - log₂ 8] + (-5/8) [log₂ 5 - log₂  8] =

(-3/8) [log₂ 3  - 3] + (-5/8) [ log₂ 5  - 3] =

-3 [ -3/8 -5/8]  - [ (3/8) log₂ 3 + (5/8)  log₂ 5 ] =

3  - [ 1/8)] * [ 3 log₂ 3 + 5 log₂ 5 ]  =

3 - [1/8] * [ log₂ 3^3 +  log₂ 5^5] =

3 - [1/8] * [ log₂ 27 + log₂ 3125] =

3 - [1/8] * [log₂ (27 * 3125)] =

3 - [log₂ 84375] / 8 =

[24 -  log₂ 84375 ] / 8 =  

 [about  0.954434]

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 0=4(e^(2x)) - e^(-x)    rearrange

e^(-x)  = 4e^2x   (1)

1 /e^(x)  = 4e^(2x)  multilpy both sides by e^x   and divide both sides by  4

1/4 = e^(3x)   take the Ln of both sides

Ln(1/4) = Ln e^(3x)   and, by a log property, we can write

Ln(1/4)  = 3x Lne    and Lne = 1  ....so we have

Ln(1/4)  = 3x     divide both sides by 3

Ln(1/4)/3  = x =  about -0.462

Here's the graph of both sides of (1) above showing the approximate solution.......