Quick calculus question work doesn't have to be shown

1 answer:

5 0

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\implies V=\cfrac{4\pi }{3}r^3
\\\\\\
\cfrac{dV}{dt}=\cfrac{4\pi }{3}\cdot \stackrel{chain~rule}{3r^2\cdot \cfrac{dr}{dt}}\quad 
\begin{cases}
\frac{dr}{dt}=50\\
r=6
\end{cases}\implies \cfrac{dV}{dt}=\cfrac{4\pi }{3}\cdot 3(6)^2\cdot 50\\\\\\ \cfrac{dV}{dt}=\stackrel{\frac{cm^3}{s}}{7200\pi}$

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Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula

Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

$2x ^ 2-4x + 5 = 0$

Where:

$a = 2\\b = -4\\c = 5$

We have to: $x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting we have:

$x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}$

By definition we have to:

$i ^ 2 = -1$

So:

$x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}$