if we take 64 to be the 100%, how much is 6¼% off of it?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%2064%26100%5C%5C%20x%266%5Cfrac%7B1%7D%7B4%7D%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B6%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B%5Cfrac%7B100%7D%7B1%7D%7D%7B%5Cfrac%7B25%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B25%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B64%7D%7Bx%7D%3D16%5Cimplies%2064%3D16x%5Cimplies%20%5Ccfrac%7B64%7D%7B16%7D%3Dx%5Cimplies%204%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bit%20had%7D%7D%7B64%7D-%5Cstackrel%7B%5Ctextit%7Bleakage%7D%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bremaining%7D%7D%7B60%7D)
Answer:
(-9, π/5 + (2n + 1)π)
Step-by-step explanation:
Adding any integer multiple of 2π to the direction argument will result in full-circle rotations, which are identities, so this family is equivalent to the give coordinates:
(9, π/5 + 2nπ), for any integer n
Also, multiplying the radius by -1 is a point reflection, equivalent to a half-turn rotation. Then add π to the direction for another half turn, and the result is another identity. So this too is equivalent to the given coordinates:
(-9, π/5 + (2n + 1)π), for any integer n
Answer:
Bet
Step-by-step explanation:
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
Answer:
(2x^3+4x^2-4x+6) / (x+3) = 2x^2-2x+2 => B
Step-by-step explanation:
Answer:
subtract 2x on both sides
Step-by-step explanation:
