This is my code for the quadratic equation. It keeps telling me that my else is illegal since no matching if statement and my else statement is missing a statement
#include <iostream>
#include <cmath>
#include <string>
#include <iomanip>
using namespace std;
int main() {
string Name;
[Code] .....
This is my code:
Code:
#include <math.h>
#include <stdio.h>
int main(void) {
float a,b,c,root_1,root_2;
printf("Please enter value a from the quadratic equation
[Code] ......
And I keep getting this error:
Code:
/tmp/ccgtUIun.o: In function `main':
assign345.c:(.text+0xc7): undefined reference to `sqrt'
assign345.c:(.text+0xef): undefined reference to `sqrt'
collect2: ld returned 1 exit status
soo quadratic equation solver. output is "x=nan x=nan".
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
//prototyping
double ans_1(double,double,double)
[code].....
I am creating a program that solves the quadratic equation ax^2 + bx +c.
I have this program almost complete except the output of the equation in the function called display_quadratic. I need the program to display the variables a,b,c in the equation ax^2 + bx + c but I am having 2 problems. My first problem is that I cannot get the right addition and subtraction signs for the equation.
For instance, if the program had the values for a,b,c as 2,2,3
it will display 2x^2 2x 3
How can I get it to display 2x^2 + 2x + 3? or if it was negative like 2x^2 - 2x - 3?
My next question is how do I get to not display the coefficients that are 1?
I had an if-else statement but no matter what I created it would overlap with another statement and print out twice. Here is the code:
Code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void display_quadratic (float a, float b, float c);
float root1(float a, float b, float c);
[Code] ......
So, I successfully made a program that will perform the quadratic equation on three numbers, imaginary or real. however, i am now trying to simplify the result, as to get rid of the "/2a" on the bottom. Hence the simplify() function. I just started to create the simplification function, and am attempting to divide the imaginary part of the solution as well as the real part of the solution by 2a. Somehow, it gives the error, "error:invalid operands of types 'int' and 'double *' to binary 'operator*'" on lines 105 and 106. I suspect it has to do with the pointers and references that i am passing as parameters. Also, just an aside, I have never actually seen "/=" be used. It can be, right? I know "+=" can be.
#include <iostream>
#include <string>
#include <cmath>
#include <cstdlib>//simplify the answer
using namespace std;
int count=0;
//prototyping
double ans_1(double,double,double);
[Code] ....
We are suppose to build a program to do the quadratic formula, which isnt really a issue. my issue is i have a if else loop and my if is being ignored.
what i want it to do is is state that if the (b * /> - 4(a)© is a negative, then it prints out stating it cant do this.
note: i know that i can use the i number system but we have been asked to not too and to do this instead.
here is my code
#include <stdio.h>
#include <math.h>
int main() {
float a, b, c, d, e, f, g, h, i; /* i would love to figure out a way to do this without so many variables*/
[code]....
I'm having an issue coming up with an if() statement to check if a word match the one in the value of a pointer's address. So far the best I've come up with only matches the first letter of the words, you'll find it in the code below.
#include"Header.h"
int Colour(struct MyStruct *ArrayPointer, int ArraySize) //ArraySize = 3 for this run. {
int ColourCount = 0;
for (int i = 0; i < ArraySize; i++) {
[Code] ....
An example run you can see in attached pic.
I want to have an if statement that only accepts "Red" and not the occasional "Ravaged_Anus".
I'm using MVS Express 2013, .c source files, and the C++ compiler.
Attached image(s)
I have a task to find errors in this long line of code in order to correctly calculate the solutions of quadratic equations.
Code:
int main(int argc, char *argv[]) {
int validInput, solution_type;
double a, b, c, root_1, root_2, q;
/***********************
* Input / Validation
***********************/
/* Check numbers of arguments, and read input */
validInput = (argc = 4);
[Code] ....
Is a section of the code (the first section). And as you can probably guess, it goes on to calculate for a > 0 etc...
I dont really understand what the validinput section is saying? And a, b and c are never defined so Xcode is just saying a,b,c,root1,root2 are uninitialized and I also dont know what means. Do I need to define these values?
Code:
int main(int argc, char *argv[]) {
int validInput, solution_type;
double a, b, c, root_1, root_2, q;
[Code] ......
I have given following exercise in my cpp book: Determine the roots of quadratic equations
ax^2 + bx + c = 0
using formula
x = -b +(plus-minus) (root)b^2 - 4ac/2a
i don't now how to write such code...
write codes that could solve a quadratic formula, and my codes are like this:
#include <bjarne/std_lib_facilities.h>
int main()
{
cout << "Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c:
";
cout << " a? ";
double a;
cin >> a;
cout << " b? ";
[code]....
Which runs perfectly, but I have 2 questions:
1. How to simplify these code? On the assignment sheet the professor wrote about using void solve_linear(double b, double c); and void solve_ quadratic (double a, double b, double c);which I currently dont understand how these works. He asked us to write a well-encapsulated (as short as possible) program.
2. These are for extra points: the precision problem of floating numbers: professor asked us to find a way to get the precise answer of it, like this: Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c:
a? 1
b? -20000
c? 1.5e-9
Trying to solve the quadratic equation 1*x*x + -20000*x + 1.5e-09 == 0
Using classical formula: Two roots, x = 20000 and x = 0
Using stable formula: Two roots, x = 20000 and x = 7.5e-14
My guess is that
A. Using code like if(x1*x2=a/c) to check if numbers were approximated.
B. Somehow determine the larger one in x1 and x2.
C. Somehow use that larger one to do something
how to take two variables with values such as a 1; and b 2;. Now you take cout << a + b; and you will get 3, but I want to somehow have a 1; b 2; and c num; then have a + b = c. So what ever I have the user input is for a and be added together will become the value of the c variable.
View 1 Replies View RelatedI have a small project for school due today but i cant get my head around programming
Ive to create a program where ive to get the values from the user and add them to an equation. I have also to add a selection statement where it will ask for the material being tested which will have its value set in the program.
Write a program that creates the output shown in the Output Layout section below. The program should create 2 points with x and y coordinates as integers, prompt the user to input the x and y values for one of the points and randomly set the other (-99 to 99 range) and output the length of the radius line segment and the area of the circle that radius defines. The program should then end. Include an SDM for the program and any other appropriate documentation.
Special Calculations:
Distance between 2 points equation:
√((p0x – p1x)2 + (p0y – p1y)2) (This requires use of the math library)
Output Layout: (bold text represents user input)
Please enter the location of your first point.
Enter a value from -99 to 99 for your x coordinate: -2
Enter a value from -99 to 99 for your y coordinate: 17
The location of your second randomly set point.
Your x coordinate: 45
Your y coordinate: -89
The length of the radius line segment from point one to point two is 115.
The area of the circle with a radius of 115 is 41546.33.
Other:
int pAx;
int pAy;
int pBx;
int pBy;
I'm writing an algorithm on a microcontroller in C.
One of the iterations that takes a long time involves the following equation, summing the product of pairs of variables:
signed char VarA;
signed char VarA;
signed long VarA;
Total+=VarA*VarB;
The vast majority of cases, at least one of the variables VarA and VarB will be 0, so nothing will be added to Total.
Is there a way to check for this condition which might give me a performance increase over doing a lot of multiplying-by-zeros?
I am having trouble understanding how this loops would work. Give the function (x*x*x*x) - (10*x*x*x) + (35*x*x) - (50*x) + 24 Write a program that will use bisection method to find the roots of this function. Define lower limit and upper limit (e.g. -1.05 and 6.05) Starting at the lower limit step along the X axis at intervals of 0.1 for H calculating the function values f(x) and f(x +h), then f(x+h) and f(x+2h) until the upper limit is exceeded. If sign of the function value changes this indicates a root between ranges. Apply bi sectional method to this range until root has been found with an epsilon of 0.000001, then continue on until upper limit has been exceeded and all 4 roots found. If the function is within 0.000001 of zero then root has been found therefore no need for bi sectional method.
Code:
#include "stdafx.h"
#include "math.h"
#define H 0.1
#define epsilon 0.000001
double F(double x);
int main(void)
[code]....
So we have a function and a range. Program requires to work along the X axis at 0.1 increments until it reaches a point where the value changes from positive to negative or negative to positive. Then apply bisection method within that range to a given accuracy then print that root. Then continue on X axis until the next change of sign is found.
How do to calculate the following equation"5+((56+8)/2)" using expression builder
Note: the equation may change "50+((26+8)/12)"
I am working on a program to find the value of the current in a coil. This value satisfies the following equation:
y'=sin(2t)-[(ey-1)/(ey+1)]
which is of the form y'=f(t,y)
I know that in order to solve this I need to use the trapezoidal method to solve a differential equation, the formula is:
yn+1=yn+.5*h(f(tn,yn)+f(tn+1,yn+1) where h=tn+1-tn
I have found examples of the standard trapezoidal method but I do not think they will work because of the difference in the formulas.
My program is returning NaN even though the value in square root function is not negative -1*b + ((sqrt(pow(b,2) * -4 *(a*c))) / 2).
x = 10;
y = -1;
z = 1;
Heres the program:
main.cpp:
#include <iostream>
#include "Quad.h"
double x,y,z; // variable x = 'a', variable y = 'b' and variable z = 'c'.
[Code] ....
I would like it return the answer and not "NaN".
I've written a program that finds the maximum of a function. I now want to change it a little so that instead of evaluating the function y=x^2-7x-18, the program will ask the user to input an equation and then evaluate that equation. I'm really not sure of how to do so.
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
int main() {
int a, b, delta, x, y;
double max= -1.8 * pow(10, 308);
[Code] .....
How do I enter the equation all at once instead of prompting for each piece separately?
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
[Code].....
I want to find the root of an equation f(x) by using the secant method [URL] .....
#include <math.h>
#include <iostream>
using namespace std;
//Define function f(x)
double f(double x, double a, double b) {
return sin((a*x)/(1+x*x))*atan(b*x)+atan(x);
[Code] ....
1) Do I have to keep the function f and RootFinderSMNew in different cpp files?
2) When compiling I get the following errors:
1>------ Build started: Project: Assignment2.1, Configuration: Debug Win32 ------
1> Find Roots Secant method.cpp
[Code] .....
What am I doing wrong ?
I am using matlab R2o11 in which the programming medium is c++. I am having these variables=
T=373 t0 376
V=0.09 to 1
P=T/V
I just need a result showing ... First it should take T=373 and solve the equation using all volume values. Then it takes another temperature and do the same.
I need an equation that determines the lowest possible cost for laying power line.
The variables are as follows:
1. The width of a river
2. The distance a factory is downstream from a station on the other side of the river.
3. The cost for laying line underwater.
4. The cost for laying line on land.
All these variables will be user input each use of program.
How to make if else code below into SWITCH STATEMENT?
cout << "Enter the temperature: ";
cin >> temp;
cout << "Enter the pressure: ";
cin >> pressure;
cout <<endl;
[Code]....
Here is my first program, I had to use a lot of if/ else statements. For some reason the last "else" (bolded) is giving me an error and not letting me run the program.
#include <iostream>
using namespace std;
const float MILES_PER_KM = 1.0 / 1.61;
const float KM_PER_MILE = 1.61;
const float FREEZING_F = 32.0;
const float FAHR_PER_CELSIUS = 9.0 / 5.0;
const float CELSIUS_PER_FAHR = 5.0 / 9.0;
const float MAX_F_TEMP = 100.0;
const float MIN_F_TEMP = 0.0;
[Code] ....
And yes, the if else's are indented properly, because there are no errors for the first 3 unit types, only for the last one ( 'F' ), and i have them all exactly in the same format.
This is the error I keep getting, but I'm not sure how to resolve it.
assignment7.cpp:11: error: no matching function for call to 'Time::display() '
Time.h:18: note: candidates are: void Time::display(int, int)
Code:
#include "Time.h"
int main() {
Time tm;
tm.set(12,53);
dt.display(); //Should display 7/4/1776
[Code] .....