CPhill

The diagonal  length  =  40√2  ft

And we need to double this since we have two diagonals  = 80√2 ft

And the perimeter of the room is  4 * 40  =  160 ft

So     80√2 + 160  ≈  273 ft  of lights needed

So....the number of strands neede  = ceiling [ 273 / 30 ]  =    10 strands

Thanks, CU!!!

Using CU's diagram.....

BF = 6    and angle FPC = 75°

tan FPC  = FC /PF

tan 75 = FC / 6

6tan 75 = FC

6 tan (45 + 30) = FC

FC  = 6 [ tan 45 + tan 30 ] / [ 1 - tan45*tan30]

FC = 6[ 1 + √3/3] / [ 1 - 1*√3/3]   

FC = 6 [ 3 +  √3] / [ 3 - √3]

FC = 6[9 + 6√3 + 3 ]  / [ 9 - 3]

FC = 6 [ 12 + 6√3 ] / [ 6]

FC = [ 12 + 6√3 ]

So....BF + FC  =  6 + [ 12 + 6√3]   =  18 + 6√3  =  BC

So   AB   =   (2/√3)BC   = (2/√3) [ 18 + 6√3]  =  36/√3  + 12 =  12√3 + 12

Good plan, ILP!!!!!

Not an admin...just a mod.....don't have the power to ban anyone....sorry  !!!!

Not as difficult as it first seems......

From O, bisect chord CD  and call the point of bisection, E

When a chord is bisected by  segment drawn from a circle's center....the angle formed is = 90 °

So...ED = 4  and OD  = 6

So OED is a right triangle with angle OED = 90  and ED a leg and OD the hypotenuse  ....so.... OE  =  sqrt[ OD^2 - ED^2 ] =  sqrt [ 6^2 - 4^2]  = sqrt [ 20]  = 2sqrt(5)  =  the height of  obtuse triangle KCD

And the base of KCD  = CD  = 8

So....the area of KCD  =

(1/2)* base * height  =

(1/2 ) * 8 * 2 sqrt (5)  =

[ 8 sqrt (5)  ]  units^2

Note that.....if you scroll down on a LuckyAce message you receive......you will see a "Block" Tab

This will prevent getting any more messages from this m***n......

CPhill

The bacteria in a jar triple every 20 seconds. After three minutes, there are 275,562 bacteria in the jar. How many were in the jar at the beginning of the experiment?

Three minutes = 180 seconds

We have this

A = A₀(3)^[ ( t / 20 )       where  A0  = the original number.....A is the number after t seconds

Filling in what we know.....

275562 = A₀ * 3^(180/20 ) 

275562 = A₀ * 3^(6)           divide both sides by 3^6

275562 / 3^6   =  A₀  =  378

The common ratio, r,   is 3

The first term = 3

So....this sum is equal to

3 [ 1 - r^*8] / [ 1 - r]  =

3 [ 1 - 3^8] / 1 - 3 ]  

3  [ -6560] / [-2]  =

3 * 3280  =

9840

Look at each row in the f(x) and g(x) columns.....where the numbers are the same we have a solution

For example...in the first row we see that  we have two zeroes

So....at that x value....we have a solution.....here it is at x = -2

Can you find the other two   ????

The problem here is to try to pick the correct graph

The key is to recognize that g(x)  = 2^x   is never below the x axis

So....the second graph is incorrect

So....it appears that the intersection points on the first graph are   x = 0   and x = 2

Does this make sense  ???