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Mathematics 7 Online
OpenStudy (itsonlycdeee):

Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree. Use your graphing calculator to verify the solution graphically. 8 cos θ + 10 tan θ = sec θ

OpenStudy (anonymous):

p = 4/3, -16/33

OpenStudy (primeralph):

@cebroski You were not supposed to give the answers.

OpenStudy (anonymous):

@primeralph I am sorry that I violated the rules.

OpenStudy (primeralph):

I know you were just trying to help.

OpenStudy (primeralph):

@itsonlycdeee Yes, but don't forget that P = A^2

OpenStudy (primeralph):

What did you get?

OpenStudy (primeralph):

You are wrong in your working. Sec theta is not -16/33 and 4/3, That's sec^2 theta. You have to find the square root first.

OpenStudy (primeralph):

@itsonlycdeee I was not yelling. You haven't said anything positive to me or the question since I started; that's why.

OpenStudy (nincompoop):

ya man, don't yell :(

OpenStudy (primeralph):

@nincompoop You know me well enough.

OpenStudy (nincompoop):

LMAO

OpenStudy (primeralph):

I'm just saying, he/she should settle down and try the problem or at least say "Thank you" for getting that far into the equation; before critiquing every step.

OpenStudy (nincompoop):

ya she should. what an ingrate!

OpenStudy (itsonlycdeee):

Thank you for how far you've gotten me. But how could someone acting in such a way deserve a thank you. I am able to say thank you, but if I'm still trying to decipher what you wrote to find the answer, it doesn't really help.

OpenStudy (nincompoop):

ya it doesn't really help because she's still clueless!

OpenStudy (nincompoop):

c'mon now!

OpenStudy (primeralph):

@itsonlycdeee I didn't act any way until you started retorting but It's fine. How far have you gotten now?

OpenStudy (primeralph):

@itsonlycdeee Fair enough. Not my loss.

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