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If p is an integer between 1000 and 1030 and if the sum of the digits of p is odd then p must be odd?
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
How do you translate The sum of 6 and a number is equal to twice the number increased by 5?
6+n = 2n+5
How do you write a equation when the sum of a positive number and several negative numbers will be positive?
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
What is the sum expression for the sum of n and 5?
The sum of n and 5 is n+ 5
C program to find sum of 'n' number using function?
void main () { int no1, sum, n; clrscr() sum = 0; n = 1; printf("\n enter the number to which sum is to be generated"); scanf("%d",&no1); while(n<=no1) { sum=sum+n; n=n+1 } printf("\n the sum of %d = %d, no1, sum"); getch (); }
How do you translate the sum of 18 and a number?
18 + n
Write a C program for sum of n numbers using pointers?
void main() { int n=0,i,*p; clrscr(); printf("enter the value of n"); scanf("%d",&n); p=&n; for(i=0;i<=n;i++) { *p=*p+i; } printf("value of the sum is %d",*p); getch(); }
If p is an integer between 1000 and 1030 and if the sum of the digits of p is odd then p must be odd?
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
Can some one translate Twice the sum of a number and 3?
2(n+3)
How do you translate The sum of 6 and a number is equal to twice the number increased by 5?
6+n = 2n+5
Translate the following phrase into an algebraic expression -the quotient of the sum of two numbers and twenty?
The quotient of two and the sum of a number and twenty
The emmiter current of an n-p-n transistor is equal to?
The sum of (base current) plus (collector current).
Translate the verbal phrase into an expression 3 less than the square of a number p?
3(n-9)
How do you write a equation when the sum of a positive number and several negative numbers will be positive?
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
How would you write out a conditions when the sum of a positive number and several negative numbers will be positive?
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
A natural number N has four factors The sum of factors excluding the number is 31 How many values can N have?
N can have any of four values: 125, 161, 209, and 221. Factors of 125 are 1, 5, 25, and 125. Factors of 161are 1, 7, 23, and 161. Factors of 209 are 1, 11, 19, and 209. Factors of 221 are 1, 13, 17, and 221.
You want the code of Discrete fourier transform in C language for your image processing program using a filter function to enhance the tiff image.?
/* Discrete Fourier Transform and Power Spectrum Calculates Power Spectrum from a Time Series Copyright 1985 Nicholas B. Tufillaro */ #include #include #define PI (3.1415926536) #define SIZE 512 double ts[SIZE], A[SIZE], B[SIZE], P[SIZE]; main() { int i, k, p, N, L; double avg, y, sum, psmax; /* read in and scale data points */ i = 0; while(scanf("%lf", &y) != EOF) { ts[i] = y/1000.0; i += 1; } /* get rid of last point and make sure # of data points is even */ if((i%2) == 0) i -= 2; else i -= 1; L = i; N = L/2; /* subtract out dc component from time series */ for(i = 0, avg = 0; i < L; ++i) { avg += ts[i]; } avg = avg/L; /* now subtract out the mean value from the time series */ for(i = 0; i < L; ++i) { ts[i] = ts[i] - avg; } /* o.k. guys, ready to do Fourier transform */ /* first do cosine series */ for(k = 0; k <= N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*cos(PI*k*p/N); } A[k] = sum/N; } /* now do sine series */ for(k = 0; k < N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*sin(PI*k*p/N); } B[k] = sum/N; } /* lastly, calculate the power spectrum */ for(i = 0; i <= N; ++i) { P[i] = sqrt(A[i]*A[i]+B[i]*B[i]); } /* find the maximum of the power spectrum to normalize */ for(i = 0, psmax = 0; i <= N; ++i) { if(P[i] > psmax) psmax = P[i]; } for(i = 0; i <= N; ++i) { P[i] = P[i]/psmax; } /* o.k., print out the results: k, P(k) */ for(k = 0; k <= N; ++k) { printf("%d %g\n", k, P[k]); } }
