Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.
If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.
If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.
However, the area x*y must be constant and = 100.
We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.
We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.
So, the difference between the max and min values of x+y = 101 - 20 = 81.
Answer:
first, subtract 17 on both sides: x²+2x-16=0
this cannot be factored, so use the quadratic formula to solve for x:
b²-4ac=2²-4(10(-16)=4+64=68
√68=2√17
so x=(-2+2√17)/2 or x=(-2-2√17)/2
x=-1+√17 or x=-1-√17
Step-by-step explanation:
Answer:
The best correct option is C
x = 1.79 to x = 3
Step-by-step explanation:
Making the sport off all little thing that crowed. They are still grouped into 7. Option C is theost appropriate.
