I need to find out What is the sum of $10 and $20

CHECK YOUR GROUND CABLE LEAVING THE BATTERY.

100=1*4*25=1*(2^2)*(5^2)
1, 2, 5, 10,20, 25, 50, 100
Choose which ones you want to list.

Keep driving it & you will end up with a big bill
to completely rebuilt the transaxle

At first sign of a trans issue you stop using it,10,20 miles
is all it takes to ruin it

Have a trans shop road test,lift test & you go along for the ride

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

Where the cells to be summed are D5 through D16.

What kind of sum would that be? If it is a sum of a certain number of values that are explicitly given (2+5.5+ 16...) just enter them one after the other separated by the operator (+).
For a sum of terms that are defined by some recursion fornula then you will need a calculator with a Computer Algebra System (TI Nspire CAS, TI 92PLUS, TI Voyage 200 PLT, Casio ClassPad, HP 40G/49 G+/50G..)
A TI 83PLUS/84PLUS may be able to handle numerical sums of terms defined recursively.

you dont give me enough information?at what speed ,10,20 30 40.as you dont tell me what is happaning and when ,im going to hazard a go. tyres pressure wheel alignment,if this fails come back to me with more info.adrian,,,,,,,,,

U2 00 EPROM error 11 EPROM error





U2 04 EEPROM serial communication error 11 counter check sum error 12 adjustment value check sum error(EEPROM) 40 CRUM communication error

U2 04 EPROM error 11 Main counter error 12 EPROM error

U2 04 EEPROM serial communication error 11 counter check sum error 12 adjustment value check sum error(EEPROM) 40 CRUM communication errorYou need the service help. It is no a distans work Good luck)

try to powercycle to reboot system//if with cable provider, adv to send signal to box (if any)//if not you need to sign up for cable providers to get digital box to access sky sports if needed

check the hold switch is off and try again

if that doesn't work plug it into your computer

#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;
}