Differentiate with respect to t Assume x = x(t)
Assume y = y(t)
Assume z = z(t)

x^2 - y^3 + z^4 = 1

Step-by-step explanation:

We want to differentiate the equation:

With respect to t, where x, y, and z are functions of t.

So:

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.