A random generated list of numbers from 0 to 4 is being used to simulate an event, with the numbers 3 and 4 representing a succe
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The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is 21 square unit.
We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate
Equation of line passing through (0,5) and (2,-2) is
y-5 = [(-2-5)/(2-0)](x-0)
=>y-5 = (-7) /2x
=>y= (-7/2x)+5 -------(eq1)
Equation of line passing through (0,5) and (5,1) is
y-5 = [(1-5) / (5-0)](x-0)
=>y-5 = (-4/5)x
=>y = (-4/5)x+5-------(eq2)
Equation of line passing from (2,-2) and (5,1) is
y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)
=>y+2=(3/3)(x-2)
=>y=x-4--------(eq3)
Now, we use definite integration to find the area between the different equation of line.
So, area enclosed between the equations is given by the
area =[(-4/5)x+5 - (-7/2)x + 5)dx +
[(-4/5)x+5 -(x-4)]dx
=>area=(7/2-4/5)x dx +
((-9/5)x+9)dx
Using properties of integration,
=>area=(27/10)x dx +
(-9/5)x+9)dx
=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)
=>area=(27/20)×(25-4) + (-9/5)×24+9×4
=>area = (27×21)/20 + (-216)/5+ 36
=>area=(567/20) - (216/5) + 36
=>area= [(567-261×4)+(36×20)]/20
=>area=[(567-864)+720]/20
=>area=423/20
=>area=21 square unit.
Hence, area of triangle is 21 square unit.
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The answer is C that is X=7 and X=-1
multiple x through the given equation to get rid of a fraction. It will be
After that bring 6X to the LHS of the equation
so that it will look like the general equation that is
Use the quadratic formula (or any other approach to find the values of x
You will arrive at