CPhill

50 = a₁ +  d(5)       →  50  = a₁ +    5d    (1)

35 = a₁ + d(10)      →  35  = a₁  + 10d     (2)

Subtract (2)  from (1)

15  =  -5d      divide both sides by -5

-3 = d

And....solving for a₁, we have

50  = a₁ - 3(5)

50 = a₁ - 15

65  = a₁

So

a₃₀  = 65  -3(29)  =   -22

Well......we obviously can't answer "several hundred".......you could upload a pic of what you're working on, or just paste a few problems on one post.......we'll see what we can do.....

1 / (a^3 * a^8)   =  a^x

1 / (a^11)   = a^x         [remember......1/a^n   = a^(-n)  ]

a^(-11)  = a^x

So   ...    x  must  =   -11

(1/2)^x  = 4

Note that 1/2  =  2^-1     and 4 = 2^2    .....so....

(2^-1)^x  = (2)^2

2^(-x)   = (2)^2           and since the bases are the same, we can solve for the exponents

-x  =  2                  multiply both sides by a negative

x = -2

Hint:  divide both sides by 5

I assume this might be :

-3x + 11  = -8x + 51    subtract 11 from both sides and add 8x to both sides

5x  = 40      divide both sides by 5

x = 8

Connect BC

Then m< CBA  = (1/2) minor arc AC   = (1/2)(164)  = 82 = m<CBP

And m< DCB  = (1/2) minor arc BD  = (1/2)(146)  = 73  = m< PCB

Then PCB  forms a triangle

And, by the Exterior Angle Theorem, m<BPD = m< CBP + m<PCB  =  82 + 73  = 155

(1/3 )(8)=(2x+1)(3x+1)

(8/3)  = 6x^2 + 5x + 1        mutiply through by 3

8  = 18x^2 + 15x  + 3      subtract 8 from both sides

18x^2 + 15x - 5  = 0      

Using the Quadratic formula, the two solutions are

x = [ - 5  ±√ 65 ] / 12

3 - 8x  ≥ 9 + 2(1 - 4x)

3 - 8x ≥ 9 + 2 - 8x        simplify and add 8x to both sides

3 ≥  11

This is not true.....so.......no value of x will make this inequality true

Good job, Rarinstraw.....!!!!