Find sin A A. Sin A=4/5

Mathematics

1 answer:

Artemon [7]3 years ago

3 0

If you are required to find A, then A=sin inverse(4/5)

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Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

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le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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You are right. In circle equation we have =r² so the correct answer will be C od D. But we have to calculate the center of a circle.

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Complete the square:

$x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2$

And we have:

$(x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\
(x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}$

So the answer is C (the same left side of equation).

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Answer:

Simple interest=$324 Total=2124

Step-by-step explanation:

Simple interest=6%

6% of 1800=108

108x3=324

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