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Thank you very much CPhill. Do you recognize this equality and where it comes from? It is this expression used to calculate the amount remaining of a radioactive substance, namely:Ao=Ao.e^kt, which I calculated to be equivalent to 2^(-t/h), where t=years and h=half-life of the radioactive substance. See here:  http://web2.0calc.com/questions/m-me-kt-m-25-t-10-how-can-i-possibly-work-out-m-if-i-don-t-know-k-the-question-asks-this-a-particular-radioactive-substance-modelled-by-equation

3^(-4)  =

1 / [ 3^4]  =

1/81

 e^(ln(1/2)*1/6*10)=2^(-10/6)

e^[ln(1/2) * ( 10/6) ] =

e^ [ (10/6) ln (1/2) ]  =

e^ [ ln (1/2)^(10/6) ] =

(1/2)^(10/6) =

2^ ^(-10/6)        [  this is true because  (1/a)^b =   a^-b ]

Good job, MW...........very well explained......  !!!!

Let x be the longer side of the rectangle and let y be the shorter side of the rectangle

And the expression for the total fencing used is

210  = x + 5y    solving for y, we have

y = [ 210 - x] / 5      this will be one side of a kennel

The other side will  =  the long side divided by 4 =  x  / 4

So....the area of one kennel =

[x /4] * [210 -x]/ 5   simplify

[210x - x^2] / 20  =

(-1/20)x^2  + 10.5x

We can find the x that maximizes this by    -b/2a     where b = 10.5 and a = -1/20

So we have

-10.5 / [ 2 (-1/20)]   =    10.5 / (1/10)   =  105 ft   ....and this is the length of the longest side

So  the length of the top side of one kennel = 1/4 of this = 105/4 ft  = 26.25 ft

And the length of the other side of one kennel  = [210 - 105] / 5   = 105/5 = 21 ft

So.....the dimensions that maximize the area of each kennel =  26.25 ft x 21 ft

opps, made a mistake here,

1/4 * 200 * 200 * π

= 10 000 π

250 000 ÷ 10 000 π ≈ 8 cm

Finding the area of the hexagon is the same as finding the area of 6 triangles of base 3 cm.

Firstly place all of the triangles in a parallelogram position

  |----------------|

        3cm

As each triangle is identical, they are all equilateral triangles, i.e. all the sides are equal.

Apply Pythagorean theorem to find out the height of the triangle.

Separate the equilateral triangle into 2 right-angled triangles.

1.5² + b² = 3² 

2.25 + b² = 9

b² = 9 - 2.25

     = 6.75

b  = 2.598076211 (10 significant figures in calculator)

Now we shift the right angled triangle like this to become a rectangle 3 × 3 cm = 9 cm length and 2.598076211 cm height.

Area of the rectangle = 9 × 2.598076211

                                  = 23.38268590 (10 significant figures in calculator)

                                  ≈ 23.4 cm²

The amount he will need is given by the whole area less the area of the windows and doors....

The whole area  = the area of the long rectangle at the bottom plus the area of the rectangle at the top left  =  (18 * 60)  + (29 * 17) = 1573 sq ft.

The area of the windows and doors =  2( 5 *3) + (2 * 15)  + (8 *4)  = 92 sq ft

So....the amount will be    =     [1573 - 92] sq ft =  1481 sq ft

radical 👌  😎 💪

small cut up tree

m=Me^kt

1/2=e^kt

1/2=e^6k

k=ln(1/2) / (6)

k=-0.11552.......

m=Me^kt

m=25e^(-.11552*10)

m=25 x .314980

m=7.87450 grams- amount remaining after 10 years.

P.S. YOU CAN USE A MUCH SIMPLER METHOD TO GET THE SAME ANSWER AS FOLLOWS:

10/6=1.666667 half-lives elapsed after 10 years.

m=25 x 2^-1.666667

m=25 x .314980

m=7.87450 grams- amount remaining after 10 years.

YOU MAY BRING IT TO THE ATTENTION OF YOUR TEACHER.