Solution: Multiply both sides by 2: $ mv^2 = \frac{3}{2}kT $. Divide by $ m $: $ v^2 = \frac{3kT}{2m} $. Take the square root: $ v = \sqrt{\frac{3kT}{2m}} $. \boxed{v = \sqrt{\dfrac{3kT}{2m}}}

February 22, 2026

$Solution: Multiply both sides by 2: $ mv^2 = \frac{3}{2}kT $. Divide by $ m $: $ v^2 = \frac{3kT}{2m} $. Take the square root: $ v = \sqrt{\frac{3kT}{2m}} $. \boxed{v = \sqrt{\dfrac{3kT}{2m}}}$

\]Question: Find the vertex of the parabola modeling the growth rate of a climate-dependent species, given by $ y = -2x^2 + 8x - 5 $.
Solution: The vertex form of a parabola $ y = ax^2 + bx + c $ has its vertex at $ x = -\frac{b}{2a} $. Here, $ a = -2 $, $ b = 8 $, so $ x = -\frac{8}{2(-2)} = 2 $. Substituting $ x = 2 $ into the equation: $ y = -2(2)^2 + 8(2) - 5 = -8 + 16 - 5 = 3 $. Thus, the vertex is at $ (2, 3) $. \boxed{(2, 3)}
Question: Solve for $ v $ in the equation $ \frac{1}{2}mv^2 = \frac{3}{4}kT $, where $ m $ is mass, $ k $ is Boltzmann’s constant, and $ T $ is temperature, to express $ v $ in terms of $ k $, $ T $, and $ m $.
Question: An ornithologist tracks a bird’s migration path, which follows the quadratic equation $ y = -x^2 + 6x - 8 $. Determine the maximum height $ y $ reached during the flight.
Solution: The maximum height occurs at the vertex. For $ y = -x^2 + 6x - 8 $, $ a = -1 $, $ b = 6 $, so $ x = -\frac{6}{2(-1)} = 3 $. Substitute $ x = 3 $: $ y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1 $. The maximum height is $ 1 $ unit. \boxed{1}
Question: Given $ f(2x + 1) = 4x^2 + 12x + 9 $, find $ f(x^2 - 3) $.

Solution: Multiply both sides by 2: $ mv^2 = \frac{3}{2}kT $. Divide by $ m $: $ v^2 = \frac{3kT}{2m} $. Take the square root: $ v = \sqrt{\frac{3kT}{2m}} $. \boxed{v = \sqrt{\dfrac{3kT}{2m}}}

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