Recent content by arroww

Oh, oops. ;p Alright, so it's pretty much the same thing when rearranging: 1 - tanh^2x = 1/cosh^2x 1 = 1/cosh^2x + tanh^2x

Oh, so rearranging it gets: 1 - tanhx = 1/cosh^2x 1 = 1/cosh^2x + tanhx

Isn't it tanhx?

It's 1.

$\dfrac{\cosh^2 x}{\cosh^2 x}$ is the RS (1). So does that mean $\dfrac{\sinh^2 x}{\cosh^2 x}$ is the LS? though I'm not sure how it is.

Hm I'm not sure. The only think I can think of is subbing $\frac{e^x + e^{-x}}{2}$ for coshx and $\frac{e^x - e^{-x}}{2}$ for sinhx.

$\frac{cosh^2 x}{cosh^2x} - \frac{sin^2x}{cosh^2x}$ so then you'd get, - $\frac{sin^2x}{cosh^2x}$ ?

Oh, so it's: $\frac{cosh^2 x - sin^2x}{cosh^2x} = \frac {1}{cosh^2x}$

$\frac{cosh^2 x - sin^2x}{cosh^2x}$ ?

That was actually the previous question. So I'm not sure what you mean?

a) Solve the trigonometric equation: A = cosX + AsinX, for some angle X. b) The imaginary number i is equal to √-1. Use part a) to solve i = cosX + isinX. You will have used degrees to answer parts a) and b). However, the mathematics below requires the use of radians. Convert degrees to...

6b) tanh^2(x) + 1/cosh^2(x) = 1 Could someone help start me off? I know that you have to sub in (e^x + e^-x)/2 for cosh and (e^x - e ^-x)/(e^x + e ^-x) for tanh. Then I'd add these together, but I'm not sure how I'd solve/simplify them arithmetically after that. Help would be appreciated! thanks.

9) A student records the internal temperature of a hot sandwich that has been left to cool on a kitchen counter. The room temperature is 19 degrees Celsius. An equation that models this situation is $$T(t) = 63(0.5)^\frac{t}{10} + 19$$ where $$T$$ is the temperature in degrees Celsius and $$t$$...

How would you find the equation for the horizontal asymptote of the following exponential function?: $$g(x) = -2f(2x - 6)$$ Let $$f(x) = 4^x$$

So, this is probably really simple...but I keep getting the wrong answer when trying to simplify this: $$3\sqrt{\frac{(10x^3)^2}{(10x^6)^{-1}}}$$Could someone show the steps to simplifying it? Thanks so much. (: