A quadratic function is a function of the form
. The
vertex,
of a quadratic function is determined by the formula:
and
; where
is the
x-coordinate of the vertex and
is the
y-coordinate of the vertex. The value of
determines if the <span>
parabola opens upward or downward; if</span>
is positive, the parabola<span> opens upward and the vertex is the
minimum value, but if </span>
is negative <span>the graph opens downward and the vertex is the
maximum value. Since the quadratic function only has one vertex, it </span><span>could not contain both a minimum vertex and a maximum vertex at the same time.</span>
Yes, ode45 can be used for higher-order differential equations. You need to convert the higher order equation to a system of first-order equations, then use ode45 on that system.
For example, if you have
... u'' + a·u' + b·u = f
you can define u1 = u, u2 = u' and now you have the system
... (u2)' + a·u2 + b·u1 = f
... (u1)' = u2
Rearranging, this is
... (u1)' = u2
... (u2)' = f - a·u2 - b·u1
ode45 is used to solve each of these. Now, you have a vector (u1, u2) instead of a scalar variable (u). A web search regarding using ode45 on higher-order differential equations can provide additional illumination, including specific examples.
Answer:
Step-by-step explanation:
9, because 18 multiplied by 3 is equal to 54
54 multiplied by 2 is equal to 108
And,
108 divided by 12 is 9.
Answer:
Area=11,309.73
Step-by-step explanation:
area of a circle is πr^2 so half of the diameter is the radius so it's 60.
pi x 60^2= 11,309.73
Answer:
ΔBFG
Step-by-step explanation:
Since top face points are
and bottom face points are
The side BF can be a hypotenuse of the triangle BFG only
This triangle is on the top face and all the rest given triangles are formed by the points that are not on the same face, therefore BF can't be a hypotenuse of those, as ether B or F is becoming a right angle of the triangles.