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2 == E x 1/2^(600/120)

2 ==E x 1/2^(5)

2 ==E  x  0.03125

E ==2 / 0.03125

E == 64 grams - amount of original element.

Here's one way of showing it:

8(5) + 8(2) + 1/2 * 3.14(8+2-4)^2/2

40 + 16 + 1/2 * 3.14 (3^2)

40 + 16 + 14.13

70.13 km^2

wheres the picture.

Let the total amount of students in the class be \(s\). 

Note that \({5 \over 9}\)th of the students will get 4 sweets, \({1 \over 3} \times (1 -{5 \over 9}) = {4 \over 27}\)th of the students will get 3 sweets, and \({8 \over 27}\)th of the students will get 2 sweets. 

From this, we can form the equation: \(({5 \over 9}s \times 4) + ({4 \over 27}s \times 3) + ({8 \over 27}s \times 2) = 264\). 

Simplifying, we get: \({88 \over 27}s = 264\), meaning there were originally \(\color{brown}\boxed{81}\) students. 

3y =  2x^2 -16x + 18 - x^2 + 12x             simplify as

3y  = x^2  - 4x +  18          divide through by 3

y =  (1/3)x^2  - (4/3)x + 6

m =  (4/3) / ( 2 *1/3)  =  (4/3) / (2/3)  =  2

n =  (1/3)^2^2  - (4/3)(2) + 6 =       (4/3) - 8/3  + 6 =   14/3

m + n =   2 + 14/3     =         20 / 3

 x==7*24*48*60==483,840 = 2^9 * 3^3 * 5 * 7

y==5^2 x  7^2 ==1,225

483,840  x  1,225 ==592,704,000

592,704,000^(1/3) == 840

3 x 11  rectangle at top =    33  cm^2

(11 - 3) x 4  =   8 x 4 rectangle in middle =  32  cm^2

2 x 2 square at bottom   =   2^2  = 4 cm^2

Total area  =  33 + 32 + 4  =  69  cm^2

Assuming that  52 is in  degrees

x =  850 * tan (52° )    ≈  1087.95

Take area of   FGIH =  (10) (11)   = 110

Subtract  areas of  the right triangles  (1/2) [ (4*5 ) + (5*2) + (6*5) + (7*4)  ]  =  44

Area of pentagon =   110   -  44  =      66

Thanks but  please show some working Vinculum. 

I have no idea how you got this.

a/b =  3/5

b /c = 6/15

c / d =  6/1

(a/ b) ( b / c)  =    (a/c) =   ( 3/5) (6/15) =  18 / 75  =    6/25

(a/c) (c/d)  =   ( a/ d) =     ( 6/25) (6/1)   =   36 / 25

Simplify as

5x^5 + 28x^3 

x^3 ( 5x^2  +  28)

Rearrange as

x^2 - 10x + 30 =  0

Sum of the roots =   10 / 1  =  10       so a =  5

Product of the  roots =  30

(5 + bi) ( 5 - bi) =   30

25 + b^2  = 30

b^2  =  30  - 25

b^2 =  5

b = sqrt (5)

ab = 5sqrt (5)

There  are    29 * 12 =    348   parts to the segment

AB = ( 5/12) *348 =   145/348

BD =  (7/12) * 348  =    203 /348

AC =  (18/29 ) * 348 =   216/ 348

CD =  ( 11/ 29 ) * 348 =  132/348

A            B         C             D

      145        71          132

AB : BC  : CD   =   145 : 71 : 132

Hello CPhill,

When calculating a statistic on indistinguishable elements or arraignments, they are (usually) treated as if they are distinguishable.  Though not distinguishable as one from another, their appearance is proportional to their multiplicity, and this affects their appearance counts in statistical measures.  

GA

--. .-

$\displaystyle \sqrt{125} \ or \ 5\sqrt{5}$ 

now can you help me with my question?

What about 5,5?

Recall the formula for probability: \(\text{succsess} \over \text{total outcomes}\).

In this case, the successes are when it is caffeine-free and fruity, and the total outcomes are the sum of all the outcomes.

Can you take it from here? 

406) We know that \(x^2 - 6x \) equals either 9 or -9. Solve for x in both cases, and plug them back in to make sure they work. 

405) First, we need to find x, so we have the equation: \(4x + 3 = -9\). Solve for x, and substitute this value into the function. 

404) We need to transform x, which we can do by substituting \(x = x - 1\). This way, it becomes \(f((x-1) + 1) = f(x)\)

So, it would now become \((x-1)^2 + 3(x-1) + 5\).