CPhill

Two computers ????

The second formula is actually  ...   a_n = a₁ + (n - 1) d

Set the formulas equal to each other

nd + a₀ = a₁ + (n - 1) d     simplify

nd +  a0 = a1 + nd - d      subtract nd from both sides

a₀  = a₁  - d      add d to both sides

a₀  +  d  = a₁    which would be technically correct in your class formula when n= 1

So......the second formula becomes

a₀ + d  + (n - 1) d   =

a₀ + d + nd - d  =

a₀ + nd       ........which is the one you used in class

I think the  main difference is that your class  uses a₀  for the first term, so that a₁ is actually the first generated term of the series  [i.e., the second term]

I believe it's more common for a₁ to be represented as the first term....thus, the ACT formula is actually preferred

The two angles shown are alternate interior angles.......and these occur between parallel lines, so they are equal....and we have......

4x - 5   = 2x + 9     add 9 to both sides. subtract 2x from each side

2x  = 14          divide both sides by 2

x = 7

So

4(7)  - 5   =  2(7) + 9 

23  = 23

So....both angles = 23°

Have a good weekend, DS.....!!!!

LOL....!!!!

Don't give him any ideas, rarinstraw......!!!!!

Thanks, Alan....

BTW....here's a tip if you get stuck like this...it isn't necessarily foolproof, but it might give you a clue.....subsitute a non-zero integer for x and see if we have equivalent expressions if we " cancel" 3x out......I'll use x = 1

So

[x^2 + 3x - 4]  / [ 3x - 7]    evaluates  to    [0] / [-4]   = 0

But

[x^2 -4] /  [-7]     evaluates to [-3] / [-7]   = 3/7

So....."cancelling" 3x   does NOT result in an equivalent expression........!!!!

Shades...I think this is supposed to be :

[tan^2 x - 1]/ tan x = tan x - cot x         working with the left side

tan^2x / tanx   -   1/tanx   =

tanx   - cotx

Not too bad, huh???

You're welcome, Shades.....note that we could have also factored  (a) as,

1/6k(k + 1)(2k + 1) + (k + 1)^2   =

(1/6)[k + 1] [ k[2k + 1]  + [k + 1] [k + 1] =

(1/6) [k + 1] [ k[2k + 1] + 6 (k +1) ]  =

(1/6) [ k + 1] [ 2k^2 + 7k + 6]  =

(1/6) [k + 1] [ k + 2 ] [ 2k + 3]

And this would have been  equal  to  the substitution in (b), as follows ::

(1 / 6) (k + 1) ( [k + 1]  + 1) (2 [k + 1] + 1)  =

(1 /6) [ k + 1] [k + 2] [2k + 3]   =

Thanks, guest......here's the way to do this with a substitution

-4x-15y=-17 and -x+5y=-13

Rearranging the second equation, we have    x = 5y + 13

Substitute this into the first equation  for y

-4[5 y + 13] - 15y  = -17    simplify

-20y - 52 - 15y  = -17

-35y  =  35     divide both sides by -35

y = -1 

And using   x = 5y + 13

x = 5(-1) + 13    =   8

  (b)  1/6k(k + 1)(2k + 1)                (a)  1/6k(k + 1)(2k + 1) + (k + 1)^2  

Expanding b with the substitution, we have

(1 / 6) (k + 1) ( [k + 1]  + 1) (2 [k + 1] + 1)  =

(1 /6) ( k + 1) (k + 2) (2k + 3)   =

(1/6) ( k^2 + 3k + 2)(2k + 3)  =

(1 / 6) (2k^3 + 6k^2 + 4k + 3k^2 + 9k + 6)  =

(1 /6) (2k^3 + 9k^2 + 13k + 6)      (1)

Expanding a, we have

1/6k(k + 1)(2k + 1) + (k + 1)^2   =

(1 / 6) (k^2 + k)(2k + 1)  + k^2 + 2k + 1  =

(1/6) ( 2k^3 + 2k^2 + k^2 + k) + k^2 + 2k + 1 =

(1/6) (2k^3 + 2k^2 + k^2  + k)   + (1/6)(6k^2 + 12k + 6)  =

(1 / 6) ( 2k^3 + 9k^2 + 13k + 6)       (2)

And since (1)  = (2).....the expressions are equivalent