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Can you write a program in c that will allow the user to input ten numbers and display the total number of positive and negative numbers entered?
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); }
What is the 4th root of -8i by De morives theorem?
Find all solutions of z4 = -8i.Recall:z = a + bi, the complex formz = |z|(cos θ + i sin θ), the polar coordinate formSo we can write:z4 = [|z|(cos θ + i sin θ)]4 = |z|4(cos 4θ + i sin 4θ); -8i = 8(0 - i), and we have|z|4(cos 4θ + i sin 4θ) = 8(0 - i).Thus, |z|4 = 8, so |z|= 81/4.The angle θ for z must satisfy cos 4θ = 0 and sin 4θ = -1.Consequently, 4θ = 3pi/2 + 2npi for an integer n, so that θ = 3pi/8 + npi/2.The different values of θ obtained where 0 ≤ θ ≤ 2pi are:n = 0, θ = 3pi/8 (1st quadrant)n = 1, θ = 3pi/8 + pi/2 = 7pi/8 (2nd quadrant)n = 2, θ = 3pi/8 + pi = 11pi/8 ( 3rd quadrant)n = 3, θ = 3pi/8 + 3pi/2 = 15pi/8 (4th quadrant))Thus the solutions of z4 = -8i are(81/4)(cos 3pi/8 + i sin 3pi8) ≈ 0.6435942529 + 1.553773974 i(81/4)(cos 7pi/8 + i sin 7pi/8) ≈ -1.553773974 + 0.6435942529 i(81/4)(cos 11pi/8 + i sin 11pi/8) ≈ -0.6435942529 - 1.553773974 i(81/4)(cos 15pi/8 + i sin 15pi/8) ≈ 1.553773974 - 0.6435942529 i
What math did Euler do?
- Euler proved that the fifth Fermat number 210 + 1 isn't prime.- He proved that all even perfect numbers can be expressed as N = 2p-1(2p-1), where p is some prime number.- He found 60 amicable numbers at a time when only 3 were known.- He was the first person to use the symbol e to represent the irrational number 2.7128...- He proved that e is irrational.- He popularized the symbol π for the irrational number 3.1415...- He introduced the symbol Σ to represent summation.- He showed that powers could be imaginary or complex numbers.- He developed the trigonometric expansion of π/2.- He invented graph theory.- He developed the Euler's constant, which is an approximation of the sum of the harmonic series.- He proved Fermat's last theorem for n = 4; i.e that x4 + y4 = z4 can't be solved with positive integers for x, y, and z.- He developed the notation f(x) for functions.- He found the geometric result known as Euler's characteristic, V + F - E where V is the number of vertices, F is the number of faces, and E is the number of edges of a polyhedra that is topologically equivalent to a sphere, which directly lead to the invention of the mathematical discipline known as topology.- He developed Euler's formula, eix = cos(x) + isin(x), where i is the imaginary number.- He invented the mathematical discipline known as the calculus of variations.- He introduced the zeta function as an infinite series, ζ(s) = Σ∞n=1(1/ns), and related it to an infinite product over all of the prime numbers pn, ζ(s) = Π∞n=1(1-1/pns)-1.
What is the difference between the price of a BMW Z4 and a Ferrari?
A very basic Z4 would be about $47,000 for a 2011 Z4. Like many other cars you can buy different "editions" of a car. In the BMW Z4 there is a Z4 sDrive30i ($47,000+), Z4 sDrive35i ($53,000+),Z4 sDrive35is ($62,000+). The same goes with the Ferrari. I can't give you the exact differences but it would be $100,000+
Is Z4 manufactured by Mercedes Benz?
no the Z4 is a coupe manufactured by BMW
How many valves does the 2013 BMW Z4 have?
The 2013 BMW Z4 has 16 valves.
How many valves does the 2006 BMW Z4 have?
The 2006 BMW Z4 has 24 valves.
How many valves does the 2007 BMW Z4 have?
The 2007 BMW Z4 has 24 valves.
How many valves does the 2012 BMW Z4 have?
The 2012 BMW Z4 has 16 valves.
How many valves does the 2008 BMW Z4 have?
The 2008 BMW Z4 has 24 valves.
How many valves does the 2011 BMW Z4 have?
The 2011 BMW Z4 has 24 valves.
How many valves does the 2010 BMW Z4 have?
The 2010 BMW Z4 has 24 valves.
How many valves does the 2014 BMW Z4 have?
The 2014 BMW Z4 has 16 valves.
How many valves does the 2003 BMW Z4 have?
The 2003 BMW Z4 has 24 valves.
How many valves does the 2005 BMW Z4 have?
The 2005 BMW Z4 has 24 valves.
