anonymous
  • anonymous
A^2+X^2= sqrtx^2+4. Solve for A WILL MEDAL AND FAN
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
you mean that there is a square root of left side, or the square root (x^2+4) is really squared.
anonymous
  • anonymous
\[A^2+x^2=\sqrt{x^2+4}\]
SolomonZelman
  • SolomonZelman
A^2 + X^2 = X^2+4 A^2=4 A=2 that would be the adjacent side (Assuming that this is the contiuation of the previous problem)

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anonymous
  • anonymous
yes sir.
SolomonZelman
  • SolomonZelman
And so the adjacent side is 2, and hypotenuse is \(\sqrt{x^2+4}\), then sec would be equivalent to?
anonymous
  • anonymous
how does A = 2?
anonymous
  • anonymous
sec is equal to cos
SolomonZelman
  • SolomonZelman
No, sec is the reciprocal of cos
anonymous
  • anonymous
oops i was reading something else
SolomonZelman
  • SolomonZelman
and A=2, is what we had from: A^2 + X^2 = X^2+4
SolomonZelman
  • SolomonZelman
secant = \(\rm Hypotenuse/Adjacent=\sqrt{x^2+4}/2\)
anonymous
  • anonymous
we are solving for "a" though. I thought we are suppose to single it out.
SolomonZelman
  • SolomonZelman
and same thing you would get from: \(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{x^2+4}{x^2+4}-\dfrac{x^2}{x^2+4}}}=}\) \(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{4}{x^2+4}}}=}\) \(\large\color{black}{ \displaystyle \frac{1}{\dfrac{2}{\sqrt{x^2+4}}}=}\) \(\large\color{black}{ \displaystyle \frac{\sqrt{x^2+4}}{2}.}\)
anonymous
  • anonymous
Can you do it where you would single out the "A"
SolomonZelman
  • SolomonZelman
I have solved for A, haven't I showed how?
SolomonZelman
  • SolomonZelman
you had that: \(\large\color{black}{ \displaystyle {\rm A}^2+x^2=\left(\sqrt{x^2+4}\right)^2 }\)
anonymous
  • anonymous
i know you solved for A. but i dont understand how "A" becomes 2
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle {\rm A}^2+x^2=\left(\sqrt{x^2+4}\right)^2 }\) \(\large\color{black}{ \displaystyle {\rm A}^2+x^2=|x^2+4| }\) assuming that x is real, \(\large\color{black}{ \displaystyle {\rm A}^2+x^2=x^2+4 }\) \(\large\color{black}{ \displaystyle {\rm A}^2=4\quad \Longleftarrow \quad {\rm A}=2 }\)
anonymous
  • anonymous
omg im so sorry i over complicated this lol. i got it now.
SolomonZelman
  • SolomonZelman
it's fine, \(....\) (And same thing you will get for the secant, if you do what I initally offered.)
anonymous
  • anonymous
ok thanks a ton

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