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what do u mean i should have no rights!!!!!

look who is talken

nobody cares what you think you are a female you should have no rights 

3.14159265359

this is not appropriate 

3.141592653.....

3.14

√[64 / 121 ]  =

√64 / √121  =

8 / 11

√64/121 = 8/11 = 0.727272727272727272727272...

so would x = i?

So, what is the question? Do you monthly payments or annual payments.....????

Compute the definite integral:
 integral_0^∞ 1/(x^2 + 8 x + 20) dx
For the integrand 1/(x^2 + 8 x + 20), complete the square:
 = integral_0^∞ 1/((x + 4)^2 + 4) dx
For the integrand 1/((x + 4)^2 + 4), substitute u = x + 4 and du = dx.
This gives a new lower bound u = 4 + 0 = 4 and upper bound u = ∞:
 = integral_4^∞ 1/(u^2 + 4) du
Factor 4 from the denominator:
 = integral_4^∞ 1/(4 (u^2/4 + 1)) du
Factor out constants:
 = 1/4 integral_4^∞ 1/(u^2/4 + 1) du
For the integrand 1/(u^2/4 + 1), substitute s = u/2 and ds = 1/2 du.
This gives a new lower bound s = 4/2 = 2 and upper bound s = ∞:
 = 1/2 integral_2^∞ 1/(s^2 + 1) ds
Apply the fundamental theorem of calculus.
The antiderivative of 1/(s^2 + 1) is tan^(-1)(s):
 = lim_(b->∞) 1/2 tan^(-1)(s) right bracketing bar _2^b
Evaluate the antiderivative at the limits and subtract.
 lim_(b->∞) 1/2 tan^(-1)(s) right bracketing bar _2^b = (lim_(b->∞) 1/2 tan^(-1)(b)) - 1/2 tan^(-1)(2):
 = (lim_(b->∞) 1/2 tan^(-1)(b)) - 1/2 tan^(-1)(2)
lim_(b->∞) 1/2 tan^(-1)(b) = π/4:
Answer: |= 1/4 (π - 2 tan^(-1)(2))

b = 4c - s^2 c = 25 s = 9

b=4*25-9^2

b=100-81

b=19

What is the slope of the line given by 8x – 24y = –48

I have to sleep, I will give the answer for 2..

You probably know that: \(\ln z =\ln|z| +i \arg(z) \)

If you try \(\ln i\):

\(\ln i\\ = \ln |i| + i \arg(i)\\ = \ln 1 + i \arctan \dfrac{1}{0}\\ =\ln 1 + i \dfrac{\pi}{2}\\ = \dfrac{i\pi}{2}\)

If you try \(\ln(1+i)\):

\(\ln(1+i)\\ =\ln|1+i| + i \arg(1+i)\\ =\ln 2 + i \arctan{1}\\ =\ln 2 + \dfrac{i\pi}{4}\\ = \ln 2 + \dfrac{\ln i}{2}\)

Therefore x = i satisfies the equation.

Good proof. 

My own proof is mentioned above.

One of the correct answers, if I did not mention that x is not a real number.

1)

Verify the following identity:
tan(cos^(-1)(θ)) = cot(sin^(-1)(θ))

Multiply numerator and denominator of sqrt(1 - θ^2)/θ by 1/sqrt(1 - θ^2):
1/(θ 1/sqrt(1 - θ^2)) = ^?(sqrt(1 - θ^2))/(θ)

1 = 1:
1/(θ 1/sqrt(1 - θ^2)) = ^?(sqrt(1 - θ^2))/(θ)

θ/sqrt(1 - θ^2) = θ/sqrt(1 - θ^2):
1/θ 1/sqrt(1 - θ^2) = ^?(sqrt(1 - θ^2))/(θ)

The left hand side and right hand side are identical:
Answer: |(identity has been verified)