[10000ダウンロード済み√] 16^(x^2 y) 16x y^2)=1 219114-16^(x^2+y)+16x+y^2)=1

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16^(x^2+y)+16x+y^2)=1- Example 14Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas(ii) y2 – 16x2 = 16The given equation is y2 – 16x2 = 16Divide whole equation by 16 (𝑦2−16𝑥2)/16 = 16/16 𝑦2/16 − 𝑥2/1 = 1The above equation i 16(xy) ^2x^2 solve please Get the answers you need, now!

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 16 / (x y ) 2 / (x – y)= 1; As you can see, this will have the same outcome as the previous, with the min/max being 16 3) y = (x16)² This is in y = (xh)² k form, but there is no k, thus the min/max k = 0 This is also true for 4) y = (x16)² Let's go back to #1 and #2 While vertex form y = (xh)² k makes finding the vertex easy, it is a lot easier to know

Incoming Term: 16^(x^2+y)+16x+y^2)=1,

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