Zero because 0 times 9 = 0
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That completely depends on the value of 'x'. Whatever 'x' happens to be at the moment, you must add to this expression an amount equal to triple-x less than 7. In other words, the expression [ 7 - 3x ] must be added.
It depends on what the first # is because you always add the opposite to the first number and then it = 0. For instance -3+3=0 see how a negative was added to the positive? or it could go like this: 6+(-6)=0
-13 is the number must you subtract from -13 to get a difference of 0
0 and 0 You cannot dived by 0 (zero)
it is the number such that 15 added to negative 15=0
Mathematically, 0 (zero). In terms of appearance, a question mark, "?"
It's not pretty (because I've never written anything in C before but it worked nonetheless) but here you go. I compiled with Microsoft Visual C++ 2010 Express with a Win32 Console Application and this is everything that was written. It worked but looks kind of bulky. Tweak to your heart's content. I wasn't 100% sure what you wanted so I made three versions that do basically the same thing with slight alterations. I learned to code in C for you... This one lists each one that was either positive or negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = x0;} if(x0>0){ z0 = x0;} if(x1<0){ y1 = x1;} if(x1>0){ z1 = x1;} if(x2<0){ y2 = x2;} if(x2>0){ z2 = x2;} if(x3<0){ y3 = x3;} if(x3>0){ z3 = x3;} if(x4<0){ y4 = x4;} if(x4>0){ z4 = x4;} if(x5<0){ y5 = x5;} if(x5>0){ z5 = x5;} if(x6<0){ y6 = x6;} if(x6>0){ z6 = x6;} if(x7<0){ y7 = x7;} if(x7>0){ z7 = x7;} if(x8<0){ y8 = x8;} if(x8>0){ z8 = x8;} if(x9<0){ y9 = x9;} if(x9>0){ z9 = x9;} printf ("\nThe negative numbers are %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf.\n", y0, y1, y2, y3, y4, y5, y6, y7, y8, y9); printf ("\nThe positive number are %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf.\n", z0, z1, z2, z3, z4, z5, z6, z7, z8, z9); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); } This one tells you how many were positive and how many were negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0, neg=0, pos=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = 1;} if(x0>0){ z0 = 1;} if(x1<0){ y1 = 1;} if(x1>0){ z1 = 1;} if(x2<0){ y2 = 1;} if(x2>0){ z2 = 1;} if(x3<0){ y3 = 1;} if(x3>0){ z3 = 1;} if(x4<0){ y4 = 1;} if(x4>0){ z4 = 1;} if(x5<0){ y5 = 1;} if(x5>0){ z5 = 1;} if(x6<0){ y6 = 1;} if(x6>0){ z6 = 1;} if(x7<0){ y7 = 1;} if(x7>0){ z7 = 1;} if(x8<0){ y8 = 1;} if(x8>0){ z8 = 1;} if(x9<0){ y9 = 1;} if(x9>0){ z9 = 1;} neg = y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9; pos = z0 + z1 + z2 + z3 + z4 + z5 + z6 + z7 + z8 + z9; printf ("\nThere are %lf negative numbers.\n", neg); printf ("\nThere are %lf positive numbers.\n", pos); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); } This one sums the positive and then sums the negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0, neg=0, pos=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = x0;} if(x0>0){ z0 = x0;} if(x1<0){ y1 = x1;} if(x1>0){ z1 = x1;} if(x2<0){ y2 = x2;} if(x2>0){ z2 = x2;} if(x3<0){ y3 = x3;} if(x3>0){ z3 = x3;} if(x4<0){ y4 = x4;} if(x4>0){ z4 = x4;} if(x5<0){ y5 = x5;} if(x5>0){ z5 = x5;} if(x6<0){ y6 = x6;} if(x6>0){ z6 = x6;} if(x7<0){ y7 = x7;} if(x7>0){ z7 = x7;} if(x8<0){ y8 = x8;} if(x8>0){ z8 = x8;} if(x9<0){ y9 = x9;} if(x9>0){ z9 = x9;} neg = y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9; pos = z0 + z1 + z2 + z3 + z4 + z5 + z6 + z7 + z8 + z9; printf ("\nThe negative numbers add up to %lf.\n", neg); printf ("\nThe positive number add up to %lf.\n", pos); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); }
That completely depends on the value of 'x'. Whatever 'x' happens to be at the moment, you must add to this expression an amount equal to triple-x less than 7. In other words, the expression [ 7 - 3x ] must be added.
A solute added to water decreases the freezing point.
The number zero! Multiplied by itself = 0 x 0 = 0 Added to itself = 0 + 0 = 0!
1000 - 610 = 390 the answer is 390 610 + 390 -------- 1000 ------- note 0 + 0= 0 9 + 1 =10 so the one is carried over to 6 + 3 + 1 = 10
it doesnt :0
0, 0 and 0
-7.2+7.2 = 0
The answer is 0
0a + 0 = aa - 0 = a
You may think that there is a unique solution to that problem, but that is not the case. Any positive number can be added to the negative version of itself to obtain a sum of zero (-589 added to 589 is zero, etc.) The simplest case, of course, is to add zero to zero, getting a total of zero.