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ok i figured out all of my 13 probs except these 3 please help me with them
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dave226

by dave226 at June 13, 2011

1. Give sec x= 2.375 and 0degrees<x<360 find all the values of x to the nearest degree 2. Find the length of the major axis of the following ellipse ((x +7)^2) /9 + ((y -2)^2) /16 3. Given the following Hyperbola find the equation of the asymptotes 9x^2 – y^2 - 90x – 4y +185 = 0Top of Form

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Dave -sec x = 2.375 can be re-written as:1/cos x = 2.375, now simply invert both sides:cos x = 1/2.375 = 0.42105cos x is positive in Quadrant I and IV.arccos(0.42105) = So, this answer makes sense, since it is in Quadrant I. The other answer is in Quadrant IV. It is a reflection of the first answer over the x-axis. It equals -But, -equals 360 -65 = So, the answers are 65 and 295 degrees.Hope that helps
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Steve204 Steve204 June 13, 2011

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2. For an ellipse, the major axis is always the bigger denominator.  Here, 16 is bigger than 9, agree.  So, the major axis is parallel to the y-axis.Using the standard form for an ellipse:16 = a^2 , where a is equal to half the length of the major axis.  Since a = 4, then double that equals 8. So, the length of the major axis is 8.
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Steve204 Steve204 June 13, 2011

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For #3, you have to complete the square twice ... once for the x variable and once for the y variable.After you complete the squares and you have the hyperbola in standard form and set equal to 1, the formula for the asymptotes are:(y - k) = ± m(x - h)where, (h,k) = center of hyperbolam = slope of the line. The slope is simply the square root of the denominator under the y-term divided by the square root of the denominator under the x-term.
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Steve204 Steve204 June 13, 2011

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