1.display your name on the monitor screen. 2.read in a distance in km and display it in km and meters eg 47.45km would convert to 47 km , 450 meters write code to solve the following equation x = squareroot of asq - bsq the values of a and b to be float read from the keyboard
Please do.
double x, y; ... x = tan (y);
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
find the program in c-pgms.blogspot.com
Exactly what do you mean by 'C program in Java'
#include#include#include#define pi 3.14void main(){float degree,tanvalue;clrscr();printf("Enter degree:\n");scanf("%f",°ree);tanvalue=tan(degree*(pi/180.0));printf("tan value of %.2f is %.2f",degree,tanvalue);getch();}
You will need to use the distributive law to solve discrete series by grouping. The distributive law is a(b + c) = ab + ac. You will be removing the common factors as you go.
Oh honey, you're throwing some trigonometry at me? Alright, buckle up. The sum of tan20tan32 plus tan32tan38 plus tan38tan20 is equal to 1. Just plug in those values and watch the magic happen. Math can be sassy too, you know.
write a program to print the series 1/12+1/22+.........+1/n2 ?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
C. K. Tan has written: 'Strategic management'
You mean you have written a program, but you don't understand it? Well, how could I explain it without seeing it?