It can be anything at all, from yoctowatts to terawatts.
You must state the power output at some modulation index
before I can calculate it for some other modulation index.
By the way . . . the answer also depends on the method of modulation,
which you have not mentioned. For example, with pure FM, the transmitted
power doesn't change, regardless of the mod-index.
The Third Index Law, also known as the Power of Zero implies that any number, except zero, when raised to the power of zero, has the numerical value of 1. a0 = 1 for a =/= 0
#include using namespace std;int main(){int numberOfElemenets = 10;double myArray[numberOfElements] = {0.0};double sum = 0;for (int index = 0; index < numberOfElements; index++){cout
A key is the name of a variable in an array ($array["key"]) and the index is the position it's at ($array = ["key" => 0], the index would be 0). Keys and indices are the same if the array is not associative though ($array = [true], the key holding the value true is named 0 and is at index 0).
Index finger is 8' 0" / 2.44m
MSK is a form of FSK in "which the waveforms used to represent a 0 and a 1 bit differ by exactly half a carrier period." This means maximum frequency deviation is 0.25 fm -> m=0.25. I presume this is called "minimum" because "this is the smallest FSK modulation index that can be chosen such that the waveforms for 0 and 1 are orthogonal."
Every non-zero number, raised to the power 0 equals 1. This follows from the index laws.
#include<stdio.h> #include<conio.h> main() { int a[100]; int n,largest,index,position; printf("enter the number of elements in the array"); scanf("%d",&n); printf("enter %d elements",n); for(index=0;index<n;index++) scanf("%d",&a[index]); largest=a[0]; position=0; for(index=1;index<n;index++) if(a[index]>largest) { largest=a[index]; position=index; } printf("largest element in the array is %d\n",largest); printf("largets element's position in the array is %d\n",position+1); getch(); }
Why is 7^0 = 1 Algebraic proof. Let 'n' be any value Let 'n be raised to the power of 'a' Hence n^a Now if we divide n^a by n^a we have n^a/n^a and this cancels down to '1' Or we can write n^(a)/n^(a) = n^(a-a) = n^(0) , hence it equals '1' Remember when the lower /denominating index is a negative power ,when raised above the division line.
100
Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.
#include<iostream> #include<iomanip> #include<time.h> void print(int a[], size_t size) { using std::cout; using std::endl; using std::setw; for(size_t index=0; index<size; ++index) cout<<setw(5)<<a[index]; cout<<endl; } int main() { srand((unsigned)time(NULL)); const size_t size=10; int a[size], b[size], c[size]; // Initialise a and b with random integers (range 1-99) for(size_t index=0; index<size; ++index) { a[index]=rand()%99+1; b[index]=rand()%99+1; } // Initialise c with products of a and b. for(size_t index=0; index<size; ++index) c[index]=a[index]*b[index]; // Calculate sum of c. int sum=0; for(size_t index=0; index<size; ++index) sum+=c[index]; // Print results. std::cout<<"Array a:\t"; print(a,size); std::cout<<"Array b:\t"; print(b,size); std::cout<<"Products:\t"; print(c,size); std::cout<<"Sum product:\t"<<sum<<std::endl; }
Very good.