Question about Computers & Internet

1/2 + 1/5 = 2/5

Posted on Apr 05, 2015

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Posted on Jan 02, 2017

sum= total ,product

Mar 30, 2017 | The Office Equipment & Supplies

The sum is the result of adding two or more things together. The things don't have to be numbers, they can be vectors, matrices, polynomials, or other things.

For the other part of your question, the sum of 838 and 2 is 840.

For the other part of your question, the sum of 838 and 2 is 840.

Jan 19, 2014 | Computers & Internet

Rectangle has two pairs of equal sides.

L=length=3 units

w=width =1 unit

**Perimeter = sum of the measures of all sides.**

2 sides with measure L: sum of lengths=**2*L**

2 sides with measure w: sum of widths=**2*w**

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

L=length=3 units

w=width =1 unit

2 sides with measure L: sum of lengths=

2 sides with measure w: sum of widths=

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

Oct 29, 2013 | Mathsoft StudyWorks! Mathematics Deluxe...

Rectangle has two pairs of equal sides.

L=length=3 units

w=width =1 unit

**Perimeter = sum of the measures of all sides.**

2 sides with measure L: sum of lengths=**2*L**

2 sides with measure w: sum of widths=**2*w**

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

L=length=3 units

w=width =1 unit

2 sides with measure L: sum of lengths=

2 sides with measure w: sum of widths=

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

Oct 29, 2013 | Computers & Internet

if(sum($D$d:$D$16)<=5000,500,sum($D$5:$D$16)*.2)

Where the cells to be summed are D5 through D16.

Where the cells to be summed are D5 through D16.

Aug 25, 2012 | Computers & Internet

This sum is equal to 40+48+56+64+72+...+3024+3032+3040+3048= 40+(48+3048)+(56+3040)+(64+3032)+(72+3024)...

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

Jul 01, 2011 | Computers & Internet

You did not provide a general (n) term. However, I figured it to be something involving n^2. Hence the sum to n terms can be of the form: xn^3 + yn^2 + zn. Using the partial sums: 1, 6 (1+5), and 18 (1+5+12) we can build a linear equations system:

1= x + y +z (n=1)

6 = 8x + 4y +2z (n=2, n^2=4, n^3 = 8)

18 = 27x + 9y + 3z (n=3, n^=9, n^=27)

Solving for x,y,z we get x = 1/2, y= 1/2, z =0

Hence 1+5+12+22+..+ f(n^2)= 1/2(n^3+n^2)

1= x + y +z (n=1)

6 = 8x + 4y +2z (n=2, n^2=4, n^3 = 8)

18 = 27x + 9y + 3z (n=3, n^=9, n^=27)

Solving for x,y,z we get x = 1/2, y= 1/2, z =0

Hence 1+5+12+22+..+ f(n^2)= 1/2(n^3+n^2)

Nov 04, 2010 | Refrigerators

Q.2# Dim n, sum as Double

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Sep 16, 2008 | Microsoft AccRepair DS2 2004 for Fr...

#include <stdio.h>

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

Feb 21, 2008 | Computers & Internet

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