anonymous
  • anonymous
Use a trigonometric ratio to find the value of x. Round the answer to the tenths’ place.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
BloomLocke367
  • BloomLocke367
Do you know about sin, cosin, and tangent and how to use them?
anonymous
  • anonymous
Yes.

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More answers

BloomLocke367
  • BloomLocke367
I was asking @barbdewysveck1 XD
anonymous
  • anonymous
no
anonymous
  • anonymous
Sorry..
BloomLocke367
  • BloomLocke367
don't be lol
anonymous
  • anonymous
I figured it out! Thank you!!!!!
BloomLocke367
  • BloomLocke367
umm... didn't @barbdewysveck1 ask this question?
anonymous
  • anonymous
whats the answer @KayKay83
dinamix
  • dinamix
cos20 = 11/x @barbdewysveck1 i help u little
BloomLocke367
  • BloomLocke367
You shouldn't give or receive direct answers as they are against the CoC. @barbdewysveck1
BloomLocke367
  • BloomLocke367
but I'll happily walk you through what to do
anonymous
  • anonymous
I got 2x^3 + x^2 + 4x - 3
anonymous
  • anonymous
k just fast please
anonymous
  • anonymous
\[2x^3+x^2+4x-3\]
anonymous
  • anonymous
wrong question @BloomLocke367
anonymous
  • anonymous
i mean @KayKay83
anonymous
  • anonymous
srry @BloomLocke367
BloomLocke367
  • BloomLocke367
well you (should) know that \(\large sin=\frac{opposite}{hypotenuse},~cos=\frac{adjacent}{hypotenuse},~tan=\frac{opposite}{adjacent}\), right?
BloomLocke367
  • BloomLocke367
we are working on the problem you posted, right?
anonymous
  • anonymous
yep
BloomLocke367
  • BloomLocke367
okay. Do you understand what I just posted?
anonymous
  • anonymous
no
anonymous
  • anonymous
No worries.
BloomLocke367
  • BloomLocke367
well, basically, that just tells you whether to use sine, cosine, or tangent to find the missing piece. You want to start from the angle, and opposite, hypotenuse, and adjacent all are referring to the placement of the angle you're looking at.
BloomLocke367
  • BloomLocke367
|dw:1441414606178:dw| That is the first triangle you have.
BloomLocke367
  • BloomLocke367
you want to be looking, first, at the angle they gave you, which is angle a. You already know the measurement now where is x in relation to a? is it the hypotenuse, is it adjacent to that angle, or is it opposite of that angle?
BloomLocke367
  • BloomLocke367
@barbdewysveck1
anonymous
  • anonymous
is the hypot the 11 or he longer one
BloomLocke367
  • BloomLocke367
the hypotenuse is always the side that is across from the right angle. so it's the longest one.
anonymous
  • anonymous
k
BloomLocke367
  • BloomLocke367
are you following me? or do I need to re-explain something?
anonymous
  • anonymous
im followin
anonymous
  • anonymous
Let f(x) = 2x^2+x-3 and g(x) = x-1. Perform the indicated operation, then find the domain. (f*g)(x)? is my answer correct? , 2x^3 + x^2 +4x -3: domain: positive real numbers? This was my question....so sorry for confusion!. Never mind. ;)
BloomLocke367
  • BloomLocke367
okay, so you know that side x is the hypotenuse.
BloomLocke367
  • BloomLocke367
the only other side you need to worry about is the one they gave you. is AC (11) adjacent to or opposite angle A?
BloomLocke367
  • BloomLocke367
@KayKay83, please do not post your question under somebody else's question. If you need help, open a question and you can tag me if you like and I'll help you to the best of my ability or you can find someone else to help.
BloomLocke367
  • BloomLocke367
@barbdewysveck1, are you still with me?
anonymous
  • anonymous
yep
BloomLocke367
  • BloomLocke367
okay. is AC opposite or adjacent to angle A?
anonymous
  • anonymous
adjacent
anonymous
  • anonymous
I apologize. New to this and trying to learn rules.. So very sorry. 1st time I have ever asked a question. Never again.
BloomLocke367
  • BloomLocke367
yep :) now look at what I told you about sin, cos, and tan and tell me which one involves the hypotenuse and adjacent side.
BloomLocke367
  • BloomLocke367
It's alright, @KayKay83 :) Welcome to OpenStudy! I'll pm you a link to the CoC.
BloomLocke367
  • BloomLocke367
@barbdewysveck1
BloomLocke367
  • BloomLocke367
@barbdewysveck1, are you still here?

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