geno3141

My argument for #20:

As x → π/2, tan(x) → ∞.

So, as x → ∞, invTan(x) → π/2.

As x → π/2, sin(x)  → 1

I will go back to my original analysis, except for the last step; it has a limit of π/2, not 1.

Ignore my answer for #18; I messed up!

My argument for #18:  

1) Since lim(x→∞) (x² + 1)/(x + 1)  = ∞,

2) the problem has the same limit as  lim(x→∞) invSec(x)

3) which has the same limit as  lim(x→∞) invCos(1/x)

4) and since  lim(x→∞) (1/x)  = 0,

5) it becomes  lim(x→0)  Cos(x)  =  1.

Any questions on any of the step?

I haven't looked at the other two yet.

infinity x infinity  =  infinity

Find the cube root of both sides:

 (k^3)^(1/3)  =  (1/216)^(1/3)

                k      =  (1)^(1/3) / (216)^(1/3)  

                k      =  1 / 6

Since you have no x-term on the right side, just simplify the right side:  f(-2)  =  5.

If your function had been  f(-2)  =  -2x + 7,

then you would have replaced x with -2 to get:  f(-2)  =  -2(-2) + 7  =  4 + 7  =  11.

24m - 135m/(-3)*3

Do multiplications and divisions before additions and subtractions (negative divided by a negative is positive):

24m + 45m*3

24m + 135m

159m

1/x  =  0.049

Multiply both sides times x:

1  =  0.049x

Divide both sides by 0.049:

1/0.049  =  x

x  =  1000/49  or  20.408...

5(0.2 + 0.8)^10                 (do what is inside parentheses)

= 5(1)^10                          (do exponents before multiplication  1^10 = 1)

= 5 x 1

= 5