CPhill

(-2-i ) / (-1 + 2i) =

-(2 + i) / (-1 + 2i) =

(2 + i ) / -(-1 + 2i) =

(2 + i) / (1 - 2i)      multiply the top and the bottom by the conjugate of 1 - 2i...i.e., (1 + 2i)......so we have

(1 + 2i)(2 + i) / [ 1 - 4i² ] =    (remember, i² = -1)

(2 + 4i + i + 2i²)/ (5) =

(5i) / 5  =  i

The amount remaining is given by

A(1/2)ⁿ   where A is the original amount and n is the number of days that have elapsed

So we have

A(1/2)²  =  A (1/4)  .....so 1/4 th of the subtance remains at the start of the 3rd day

They might have been a little more "cocky" than you are.....especially Zegroes....!!!!!

We all miss Zegroes......DragonSlayer554, too.....you'd like both of them, h7......!!!!

I assume we have either

log₁₀₀₀ a = 55   ... or....  log₁₀₀₀55  = a

In the first case "a"   = 1000⁵⁵  (this is a really "big" number)

In the second case, "a"  can be found by using the "change-of-base" rule

a =   log 55 / log 1000   =  log 55 / 3

The final result is given by

2(.60P)   .......where P is the original price

(Note, "40% off" means that the original price is just 60% of what it was = .60P.......and we're doubling that result)

1.  (y^2 + 2y – 35) ÷ (y+7) =    factor the numerator

(y + 7) (y - 5) / (y + 7) =          "cancel'  on top and bottom

(y - 5)

2. a^8 – 5a^5 – 3a^3 / a^2  =     factor the numerator

a^3 (a^5 - 5a^2 - 3) / a^2   =      "cancel" on top and bottom

a(a^5 - 5a^2 - 3)

-9(-7)-16y=-21    I assume you'd like to solve this

63 - 16y  = -21    subtract 63 from both sides

-16y = -84          divide both sides by -16

y = -84 / -16   =   21 / 4

Correct - a - mundo  !!!

I assume we have

1/[1 - sinΘ ] + 1 / [1 - sin Θ]  =

2 / [ 1 - sinΘ ]