Answer:
There are many factors that cause aggregate demand to shift from AD to AD1. The unemployment rate will fall and inflation will increase.
<u>Explanation:</u>
A Shift in aggregate demand from AD to AD1 means there has been a fall in demand. Various factors that cause demand to decrease are:
- Increase in price of a good itself
- Increase in the price of complimentary goods-This will lead to a fall in demand. Like ink and pen are complementary goods. if the Price of ink increases then demand for pen will decrease.
- A Decrease in the price of substitute goods-Like tea and coffee.
- Expectation regarding future fall in price
So due to the decrease in demand finally the unemployment will increase and with that, the inflation rate will increase making things dearer.
Answer:
Explanation:
The following code is written in Java. It is hard to fully create the code without the rest of the needed code including the T class and the Measurable interface. Regardless the following code can be implemented if you have that code available.
public static T minmax(ArrayList<T> mylist) {
T min = new T();
T max = new T();
for (int x = 0; x < mylist.size(); x++) {
if (mylist.get(x) > max) {
max = mylist.get(x);
} else if (mylist.get(x) < min) {
min = mylist.get(x);
}
}
return (min, max);
}
You never said what language so I used C
#include <stdio.h>
<span>int main () { </span>
<span> int a; </span>
<span> for( a = 3; a > 0; a-- ){ </span>
<span> printf("%i \n", a); }</span>
<span> printf(" Blast OFF !!!\n"); </span>
<span>return 0; </span>
<span>}</span>
Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
I really think it A : pager
I hope this helps !!
