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2 years ago

1. Triangle ABC with coordinate points A(-2,3); B(-1,0); C(5,-4) is rotated 90 degrees clockwise. What

Mathematics

1 answer:

Reil [10]2 years ago

6 0

Answer: A'(3,2), B'(0,1) and C'(-4, -5).

Step-by-step explanation:

When we rotate (a,b) 90 degrees clockwise, the new point will be (b,-a).

i.e. rotation rule for 90 degrees clockwise : (x,y) → (y,-x)

Given: Triangle ABC with coordinate points A(-2,3); B(-1,0); C(5,-4) is rotated 90 degrees clockwise.

Then, the new image points for triangle A'B'C' will be :

A(-2,3) → A'(3,-(-2)) =A'(3,2)

B(-1,0)→ B'(0,-(-1)) = B'(0,1)

C(5,-4) → C'(-4, -5)

Hence, the new image points for triangle A'B'C' are A'(3,2), B'(0,1) and C'(-4, -5).

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This code or is program to find a given value of derivative of a polynomial.

Step-by-step explanation:

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// libraries

#include <stdio.h>

#include <conio.h>

//use to control floating elements

float poly(float a[], int, float);

//main

int main()

{

// Enter the degree of polynomial equation

float x, a[10], y1;

int deg, i;

printf("Enter the degree of polynomial equation: ");

scanf("%d", &deg);

printf("Ehter the value of x for which the equation is to be evaluated: ");

// Enter the coefficient of x to the power

scanf("%f", &x);

for(i=0; i<=deg; i++)

{

printf("Enter the coefficient of x to the power %d: ",i);

scanf("%f",&a[i]);

}

// The value of polynomial equation for the value of x

y1 = poly(a, deg, x);

printf("The value of polynomial equation for the value of x = %.2f is: %.2f",x,y1);

return 0;

}

/* function for finding the value of polynomial at some value of x */

float poly(float a[], int deg, float x)

{

float p;

int i;

p = a[deg];

for(i=deg;i>=1;i--)

{

p = (a[i-1] + x*p);

}

return p;

}

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