Answer:
$60 dollars
Step-by-step explanation:
x axis is hours spent and y axis is $3 + dollars earned so 15*3 =45 +15 = 60
A: Gradient of f
B: Gradient of f at point P:
Just put the coordinates of p in above formula:
C: The directional derivative of f and P in direction of v:
The directional derivative is found by dot product of :
D: The maximum rate of change of f at P is calculated by evaluating the magnitude of gradient vector at P:
E: The (unit) direction vector in which the maximum rate of change occurs at P is:
That vector v is the needed unit vector in this case.
we divided by to make that vector as of unit length.
Learn more about vectors here:
Answer:
a) The gradient of a function is the vector of partial derivatives. Then
b) It's enough evaluate P in the gradient.
c) The directional derivative of f at P in direction of V is the dot produtc of and v.
d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.
e) The maximum rate of change occurs in the direction of the gradient. Then
is the direction vector in which the maximum rate of change occurs at P.
Sorry this is so late.
The answer is "Add the left side of equation 2 to the left side of equation 1"
Answer:
bro!! I'm on the same question on ap3x!!
lol
Step-by-step explanation:
Answer:
The minimum value of the given function is f(0) = 0
Step-by-step explanation:
Explanation:-
Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.
i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.
ii) find first derivative and equating zero
iii) solve and find 'x' values
iv) Find second derivative then find the minimum value at x=a
v) Find second derivative then find the maximum value at x=a
Problem:-
Given function is f(x) = log ( x^2 +1)
step1:- find first derivative and equating zero
……………(1)
the point is x=0
step2:-
Again differentiating with respective to 'x', we get
on simplification , we get
put x= 0 we get > 0
then find the minimum value at x=0
Final answer:-
The minimum value of the given function is f(0) = 0
END OF THE IPOD ERA
players are replacing the iPod, along with the category of device it helped to create. Sales of the iPod worldwide from 2007
through 2011 (in millions) were
approximately
N(0= -165t2 + 13.13t+ 39.9 (0 < t< 4)
in year t, where t= 0 corresponds to 2007. Show that the worldwide sales of the iPod peaked sometime in 2009. What was the approximate largest number of iPods sold worldwide from 2007 through 2011?
Answer:
a. t = 2.48 will be a period within 2009.
b. 56.16 million
Step-by-step explanation:
Here is the complete question
Apple introduced the first iPod in October 2001. Sales of the portable music player grew slowly in the early years but began to grow rapidly after 2005. But the iPod era is coming to a close. Smartphones with music and video
Apple introduced the first iPod in October 2001. Sales of the portable music player grew slowly in the
END OF THE IPOD ERA
players are replacing the iPod, along with the category of device it helped to create. Sales of the iPod worldwide from 2007
through 2011 (in millions) were
approximately
N(0= -2.65t2 + 13.13t+ 39.9 (0 < t< 4)
in year t, where t= 0 corresponds to 2007. Show that the worldwide sales of the iPod peaked sometime in 2009. What was the approximate largest number of iPods sold worldwide from 2007 through 2011?
a. Show that the worldwide sales of the iPod peaked sometime in 2009
N(t) = -2.65t² + 13.13t + 39.9
To find the maximum value of N(t), we find dN(t)/dt and equate it to zero
dN(t)/dt = d[-2.65t² + 13.13t + 39.9]/dt
dN(t)/dt = -5.3t + 13.13 = 0
-5.3t = - 13.13
t = -13.13/(-5.3)
t = 2.477
t ≅ 2.48
d²N(t)/dt² =d[-5.3t + 13.13]/dt = -5.3 < 0. So, t = 2.48 is a maximum point
Since t = 2 is 2009 and t = 3 is 2010, t = 2.48 will be a period within 2009.
b. What was the approximate largest number of iPods sold worldwide from 2007 through 2011?
The approximate largest number of ipods sold is when t = 2.48
N(2.48) = -2.65(2.48)² + 13.13(2.48) + 39.9
N(2.48) = -16.29856 + 32.5624 + 39.9
N(2.48) = 56.16384
N(2.48) ≅ 56.16 million
Answer:
The algorithm is given below.
#include <iostream>
#include <vector>
#include <utility>
#include <algorithm>
using namespace std;
const int MAX = 1e4 + 5;
int id[MAX], nodes, edges;
pair <long long, pair<int, int> > p[MAX];
void initialize()
{
for(int i = 0;i < MAX;++i)
id[i] = i;
}
int root(int x)
{
while(id[x] != x)
{
id[x] = id[id[x]];
x = id[x];
}
return x;
}
void union1(int x, int y)
{
int p = root(x);
int q = root(y);
id[p] = id[q];
}
long long kruskal(pair<long long, pair<int, int> > p[])
{
int x, y;
long long cost, minimumCost = 0;
for(int i = 0;i < edges;++i)
{
// Selecting edges one by one in increasing order from the beginning
x = p[i].second.first;
y = p[i].second.second;
cost = p[i].first;
// Check if the selected edge is creating a cycle or not
if(root(x) != root(y))
{
minimumCost += cost;
union1(x, y);
}
}
return minimumCost;
}
int main()
{
int x, y;
long long weight, cost, minimumCost;
initialize();
cin >> nodes >> edges;
for(int i = 0;i < edges;++i)
{
cin >> x >> y >> weight;
p[i] = make_pair(weight, make_pair(x, y));
}
// Sort the edges in the ascending order
sort(p, p + edges);
minimumCost = kruskal(p);
cout << minimumCost << endl;
return 0;
}