CPhill

2ab-14a+5b-35  =

(2a + 5) (b - 7)

xy+3x-4y-12  =

(x - 4)(y  + 3)

There are 3 complex numbers a+bi, c+di,e+fi. If b=3, e=-a-c, and the sum of the numbers is 2i, find d+f.

[a + bi ] + [ c + di] + [ e + fi]  =   2i

Therefore  

a + c + e  = 0  →   a + c + [-a -c]  = 0    which is true

And

b + d + f   = 2    and if b = 3, then

3 + (d + f)   = 2       subtract 3 from both sides

(d + f)   = - 1

Let the width = W

And the length = W + 4

And l x w  = area     ....so.......

(w + 4) (W)   = 60     simpliify

W^2 + 4W - 60   = 0      factor

(W + 10) (W - 6)  =  0      and setting each factor to 0, W = -10 cm [reject],    or W  = 6cm

And the length is 4 more than this =  10cm

f(x) = 17/x

Write this as 

y = 17/x      multiply both sides by x and divide both sides by y

x = 17/y     "exchange"  x and y

y = 17/x      for y, write f^-1(x)

f^-1(x)  = 17/ x         thus, amazingly..........this function is also its own inverse !!!

Let the width  = w

Let the length = 2w + 4

So.....

l * w  = 48

Substituting, we have

(2w + 4) (w)  = 48       simplify

2w^2 + 4w   = 48       subtract 48 from both sides

2w^2 + 4w  - 48  =  0      divide through by 2

w^2 + 2w  -  24  =  0          factor

(w + 6) (w - 4)   = 0       set both factors to 0   and w = -6 feet [ not possible ]  or w = 4ft

And the length, l,    = 2w + 4  =  2(4) + 4  = 8 + 4   = 12 ft

Note the reason for not be able to take a negative log.......

If we had

log (-2)     ......this would imply that  there is some "x" such that

10^x  = -2

But notice that 10 raised to  any  "real" power is always positive....therefore.......there is no "real" power "x"   that we can raise 10 to that will result in a negative.......

At the end of the 6th year [= the beginning of the 7th ], there were 449 students

At the end of the 12th year [ = the beginning of the 13th] , there were 750 students

So we have these two equations where a₁ is the beginning enrollment [in the first year]   and d is the yearly [constant ] increase

449 = a₁  + d(6 -1)            →  449 = a₁ + 5d        (1)

750  = a₁ + d(13 - 1)         →  750  = a₁  + 12d     (2)

Subtract (1)  from (2)

301  = 7d        divide both sides by 7

43  = d    this is the yearly increase

And using (1)   we can find the beginning enrollment, a₁

449 = a₁  + 5(43)

449 = a₁ + 215    subtract 215 from each side

234 = a1

So.......the equation giving the number of students for any year,  n  =

a_n   =  234 + 43 (n - 1)

logx+log(x+1)=log2    

Remember that if we have log a + log b, we can write this as log [a * b]........so....

log x + log (x + 1)   =  log [ x (x + 1) ]  =  log [x ^2 + x]

So

log [ x^2 + x  ]    = log 2         we can just forget the logs and solve this equation :

x^2 + x  = 2      subtract 2 from both sides

x^2 + x  - 2  = 0         factor

(x -1) (x + 2)   = 0       and setting each factor to 0, x = 1  or x = -2

Since we can't take the log of a negative number, x = - 2 is not a solution

Check  x = 1

log (1) + log (1 + 1) =  log 2 ?

0 + log 2   = log 2

log 2 = log 2       OK!!!

Here is [roughly] Euclid's proof of this......

Let the greatest known prime = P

Now, multiply all the known primes together [including P ] and let this number  = M

Now....add 1 to M.......so.....the new number is M + 1

If we divide M + 1 by any of the known primes, we get a remainder .......so...........either M + 1 itself is prime, and therefore greater than the greatest prime, P,  or there exists a prime greater than any of the knowm primes which divides  M and which is also greater than P

Therefore.......the list of primes is infinite