The function yp(t)=ln(3 2t), t>−32, is a particular solution to the differential equation y′′ 7y=g(t). find g(t)?

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

Step-by-step explanation:

x-intercepts are when the curve intercepts the x-axis, so when y =0.

Therefore, to find the x-intercepts, substitute y = 0 and solve for x.

The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.

Equation: $y=x(x-2)$

$\implies x(x-2)=0$

$\implies x=0$

$\implies (x-2)=0 \implies x=2$

Therefore, the x-intercepts are x = 0 and x = 2

The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1

Equation: $y=(x-4)(x+5)$

$\implies (x-4)(x+5)=0$

$\implies (x-4)=0 \implies x=4$

$\implies (x+5)=0 \implies x=-5$

Therefore, the x-intercepts are x = -5 and x = 4

The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5

Equation: $y=-5x(3-x)$

$\implies -5x(3-x)=0$

$\implies -5x=0 \implies x=0$

$\implies (3-x)=0 \implies x=3$

Therefore, the x-intercepts are x = 0 and x = 3

The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

4 0

3 years ago

Simplify 6 to the fifth power over 7 to the third power answer to the second power .

liq [111]

I think it may be (6/7)^2

3 0

4 years ago