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Calculus 1 Answer Steve M This is plotted using AutoGraph Answer link Related questions How do I determine the molecular shape of a molecule?Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel

X^3+y^3+z^3=1 graph

X^3+y^3+z^3=1 graph-Stack Exchange Network Stack Exchange network consists of 180 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share theirBecause maxima/minima occur when f ′(x) = 0, we take the implicit derivative and set it equal to 0 We should end up with a few x3y313xy Final result x3 3xy y3 1 Step by step solution Step 1 Checking for a perfect cube 11 x33xyy31

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In this math video lesson I show the student how to graph the equation xy=3 This equation is in standard form and I covert that to slope intercept form toIt's a fairly simple (but very laborious) question of exploiting divisibility With all primes > 2 out of the way, their multiples soon follow, so all that's left is n = 4 You inevitably arrive at a point where the product of two terms equals a product of n terms For n = 2 this works, but not for n = p > 2, due to coprimality – LucianX^3 x y^3 y = z^3 z (1) where x, y, z are integers greater than 1 If z and x are both odd or both even, we can define integers u and v such that z=uv and x=uv Substituting into equation (1) gives y^3 y = 2v(3u^2 v^2 1)

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 Ex 63, 1 Solve the following system of inequalities graphically x ≥ 3, y ≥ 2 First we solve x ≥ 3 Lets first draw graph of x = 3 At x = 3, y can have any value Drawing graph Checking for (0,0) Putting x = 0, y = 0 x ≥ 3 0 ≥ 3 which is false Hence, Origin does not lie in plane x ≥ 3 So, we shade right side of line Now we solve y ≥ 2 Lets first draw graph of y = 2 At y = 2, x canX^ {2}2xy^ {2}2yz^ {2}2z3=0 x 2 2 x y 2 − 2 y z 2 2 z 3 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and y^ {2}z^ {2}2y2z3 for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is

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