$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}$

5 0

3 years ago

Let g be the function given by g(x)=x^2*e^(kx) , where k is a constant. For what value of k does g have a critical point at x=2/

ArbitrLikvidat [17]

G `( x ) = $2x * e^{kx} + k x^{2} *e^{kx} = \\ = xe^{kx}(2 + kx )$
2 + k x = 0
k x = -2
k = -2: x = - 2 : 2/3 = - 2 * 3/2
k = - 3
Answer: for k= - 3, the function g ( x ) have a critical point at x = 2/3.

8 0

3 years ago