CPhill

sec^2-1/sec^2=sin^2x

tan^2x / sec^2x =

[sin^2x / cos^2x]  *  1 / sec^2x =

[sin^2x / cos^2x]  *  cos^2x

sin^2x

For the second one, we have

2. 5x³ + 15x²; x² – 6x + 9; 2x² – 6x

Factor each

5x^2(x + 3), (x -3)^2, 2x(x -3)

And the LCM is

10x^2(x + 3)(x -3)^2 =  {you could expand this...if you want}

10x^2 (x ^2 - 9) ( x- 3) =

10x^2 (x^3 - 3x^2 - 9x + 27) =

10x^5 - 30x^4 - 90x^3 + 270x^2

3.   -5/8 - 3/[2y]

Note that the common denominator is just 8y....so we have

[-5(y)  - 3(4)] / [2y] =

-[5y + 12] / [2y]

2.  [a - 1]/[a  -2] - [3a -1] / [a^2 + 3a - 10]   factor, where necessary

[a-1] / [a - 2] - [3a - 1 ] / [(a + 5)(a -2)]       notice that we need (a + 5) in the first fraction

[(a - 1)(a + 5) - (3a -1) / [(a + 5)(a -2)] =

[a^2 + 4a - 5 - 3a + 1]/ [(a + 5)(a - 2)] =

[a^2 + a - 4] / [(a + 5)(a -2)]

1. [4w + 8] / [w^2 - w - 6] * [w^3 - 3w^2] / [8w^2 - 8w]    factor

4[w + 2] / [(w -3)(w + 2)] * [w^2 (w -1)] / [8w(w -1)] =

4/(w - 3) * (w^2)/(8w)  =

w / [2(w -3)]

The answer is given by the ratio of the square of their radiuses

So we have

(300)^2 /(1.5)^2  =  40,000 times as large

Nice, dogeminer....!!!

BTW......  -0/ 0   = 0 / 0  ......and that's undefined.....

f (x) = -2x+5÷8

Write f(x) as y

y = -2x + (5/8)     subtract (5/8) from both sides

y - (5/8)  = -2x     multiply both sides by -1/2

-(1/2)y + 5/16  = x   get a common denominator on the left 

[-8y + 5] / 16 = x     "exchange"  x and y

[-8x + 5] / 16  = y     for "y"  write f^-1(x)

f^-1(x) = [-8x + 5]/ 16

Check...putting 0 into the original function leads to 5/8....and putting 5/8 into the inverse leads to 0

1. 5/b² , 6/ab

The LCM is ab^2

So we have

5a/ab^2 ,  6b / ab^2

2. 4/(n-1)² , 3/ [n(n-1)]

The LCM is n(n-1)^2

So we have

4n / [n(n-1)^2] , 3(n-1)/ [n(n-1)^2]

As an  addendum to geno's explanation.....suppose we don't want to choose any books from the shelf.....in other words...C(5,0)

The "formula" is

5! / [(5 - 0)! * 0!] = [5!/ (5! *  0!)]  = [1 / 0! ]

Well.....there 's only one way not to choose any books from the shelf.....don't take any of them!!!

Thus......the 0! in the "formula" MUST = 1  for this to "work"

OK....try to work them yourself, first......if you get stuck ..... post them....and I'll take a look....

[x^2 -25] / [x^2 - 10x + 25] * [x^2 - 7x + 10] / [x^2 - x -2] + 1/[x + 1]

Factor and "cancel" where possible

[(x + 5)(x -5)]/ [(x -5)(x -5)] * [(x -5)(x - 2)]/[ (x -2)(x + 1)]  + 1 /[x + 1]

(x + 5)/(x-5) * (x -5)/(x + 1)  + 1 /(x + 1)

(x + 5)/(x + 1)  + 1/(x + 1) =

(x + 5 + 1) /(x + 1) =

(x + 6) / (x + 1)

{Remember.....the inclination is to want to combine the last two fractions, first......however, this is incorrect......the multiplied fractions are done before any adding/subtracting takes place....... }