CPhill

2N= Ne^.10t       where N is the original population

divide by N on both sides

2 = e^.10t         take the ln of both sides

ln2 = lne^.10t    and by a log property we can write

ln2 = (.10t) lne        and lne = 1   ....so....

ln2 = .10t            divide both sides by .10

(ln2 / .10) = t ≈ 6.93 hrs.

(1-cosx)(cscx+cotx)

Well..it could be alot of things....let's see

(1-cosx) (1/sinx + cosx/sinx)

(1-cosx)(1 + cosx)/sinx

(1-cosx)² / sinx

(sinx)² / sinx

(sinx)

And there you go......

I think i'll give him thumbs-up for a good attempt !!!....those minus signs are tricky !!

Heck...I hope I got it right !!

(.6ab - .5bc + cd) - ( - .5ab + 2.5bc - cd)    ........distribute the negative across the terms inside the second set of parenthesis

.6ab - .5bc + cd  + .5ab - 2.5bc + cd        .....combine like terms....

1.1ab - 3bc + 2cd

4cmx4cmx5cm=

(16 * 5)  cm ³ =

80 cm ³

(1 - √10 )² =

(1 - √10 ) (1 - √10 ) =

1 -1√10 - 1√10 + √10 * √10  =

1 - 2√10 + 10  =

11 - 2√10

Only if  n = 1

Tan K = 2.592

Use the tangent inverse to find this

atan(2.592) = 68.9°  .......note that this could also be a 68.9° angle in the third quadrant = 248.9°  

There's an easy way to always compare two fractions

1.1 / 9.5   = 1 /20       

Multiply the denominator of the second fraction  by the numerator of the first fraction and put this number  on the "left" 

Then, multiply the denominator of the first fraction times the numerator of the second fraction and put this on the "right"

So we have

(20)*(1.1)        (9.5)*(1)   

22                      9.5  

Which side is larger ??     Then, that's the side that was greater in the first place !!

So

1.1 / 9.5   >   1 /20   

(3x²y) - (2xy²) + (10xy) - (15y) for x= -5, y=3

(3(-5)²(3)) - (2(-5)(3)²) + (10(-5)(3)) - (15(3)) =

9*25   + 90   - 150 - 45   =

225 -150  + 45   = 

75 + 45  = 

120