Mathematics
Mathematics, 28.07.2019 13:30, QueenNerdy889

What is the inverse of f if f (x) = ^3 sqrt x-7

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Mathematics, 01.08.2019 00:00, jadfavors4321
Can you me simplify these double radicals? 1.[tex] \sqrt{6+2 \sqrt{5} } [/tex] 2.[tex] \sqrt{6+2 \sqrt{8} } [/tex] 3.[tex] \sqrt{10+2 \sqrt{21} } [/tex] 4.[tex] \sqrt{7-2 \sqrt{10} } [/tex] 5.[tex] \sqrt{10-2 \sqrt{21} } [/tex] 6.[tex] \sqrt{6-\sqrt{20} } [/tex] 7.[tex] \sqrt{7-\sqrt{48} } [/tex] 8.[tex] \sqrt{8+4\sqrt{3} } [/tex] 9.[tex] \sqrt{12-6\sqrt{3} } [/tex]10.[tex] \sqrt{19+8\sqrt{3} } [/tex]
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Mathematics, 01.08.2019 20:40, mp515
So, for an investigation i'm doing, i had to take the quadratic formula and take the inverse function in order to swap the x's and y's. i did all that as seen below: [tex]y=a(x-h) x^{2} +k[/tex] [tex]y=a(x-h)^{2} +k y-k=a(x-h)^{2} (y-k)/a=(x-h)^{2} \sqrt{(y-k)/a} =x-h x=+/- \sqrt{(y-k)/a} +h therefore: y=+/- \sqrt{(x-k)/a} +h [/tex] however, i have been told that i have to explain that since i swapped the x and y axes, the x-axis will really be the y-axis when i rotate the piece (volume of rotation). and i have to explain that dx is really dy because the xs and ys are swapped in order for the program to function properly. i just have no clue how to explain it. does this make any sense: in order to make the parabola open to the right, i had to inverse the parabolic equation, resulting in the equation [tex] x=+/- \sqrt{(y-k)/a} +h [/tex] which maps a sideways parabola. however, the program that i was using, sketchpad, would not process the formula written in this way and therefore, in order for the program to function properly with the inverse function, i swapped the xs and ys, resulting in the following equation: [tex] y=+/- \sqrt{(x-k)/a} +h [/tex] . this means that when i take the integral of this function, the dx is really a dy and when graphing, the x-axis is really representative of the y-axis.
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