Recent content by mathdrama

Oh, I understand now. Thank you very much.

The amplitude is never negative? I must be doing this wrong. I had assumed the amplitude of this function: - 2 cos 3(\theta + 90°) + 1 must be -2

How exactly do you describe transformations? For instance, if the amplitude has gone from 1 to -2, how would you word that?

Are attachments acceptable or would LaTex still be more convenient? Could you kindly help me check my work?

Would this be an okay way to go about solving the problem? Also, how do I use Latex?

Oh, thank you. I finally understand now.

Okay, but I don’t how to simplify that any further or even turn it into cosθ.

Not really sure how to do this problem. I'm not even sure where the angles are. 5. The balcony of a cruise ship is 25m above sea level. A person standing on the balcony sees two buoy’s in the water below. The first buoy is situated directly east of her at an angle of depression of 32°. The...

Is it something like 1 - sin^2 = 1 = sin^2?

I don't know how to apply a Pythagorean identity, can you help me?

I have no idea how to go about proving this trig identiy. I mean, I've been taught that it's a safe bet to convert everything to sines and cosines, but other than that, I've no clue. Am I even on the right path?

Basically need more help checking my answers, I think it's better if I provide the full context: A radioactive substance with an initial mass of 300mg has a half-life of 2 years. a) Write an equation to model the mass of the material over time. Let A = mass of the material Let t = time A =...

I had to evaluate this: [32x^5y^4)^2/5/(4x^2)^2y^3/5]^-1 This is how I did it, but is it correct? = [32^2/5x^2y^4(2/5)-1 (4x^2)^(2y^3/5) = [32^2/5x^2y^8/5)^-1 (4x2)^(2y^3/5) = [(^5√32)^2*x^2y^8/5)-1 (4x2)^(2y^3/5) = [(2)^2*x^2y^8/5)-1 (4x^2)^(2y^3/5)...

I was to evaluate this: (-5x^2y^3)^-2 This is what I came up with:= 1/(-5x2y3) (-5x2y3) =1/25x4y6 Is this correct?

Re: word problem So to make sure I have this right... Let t = 0 h = -5(0^2) + 5(0) + 210 h = 0 + 0 + 210 h = 210 Therefore, the cliff is 210 meters high.