Hello,
Let me try out the solution for you.
Consider the below scenarios for the equation y = |x+5|-|x-5|
Case 1:
when x more than or equal to 5
then y=(x+5)-(x-5) = 10
hence y=10
Case 2:
when -5<x<5
y=(x+5)-(-(x-5)) = 2x
y=2x so y can take 9 values corresponding to x={-4,-3,-2,-1,0,1,2,3,4}
Case 3:
when x less than or equal to -5
y= -(x+5)-(-(x-5))
y=-10
Hence if we combine all 3 cases we get that y can take total of 11 values.
True, they all have different values !
Use the equation a^2 + b^2 = c^2.
225 + 1296 = c^2
1596 = c^2
Square root both
39 = c
Multiply the base by the height and then divide it by 2.
A={x|x is an even while number between 0 and 2} = ∅ since there is no number between 0 and 2 that is an even whole number. So there is no number to be substituted for x, resulting in an empty set.
Hope that helps!
