Answer:B? l hope is B l hope helps
Step-by-step explanation:
if not help then l will answer other problem from you l hope is right
Answer:
y=(x-6)^2+2 Have a nice day
Answer:
Quadrant 1
Step-by-step explanation:
how I understood this is first understanding how a graph works. Once you understand that you can count 13 horizontal on the "X" Axis, then go 18 up on the "Y" axis. you should get a point in the upper left area and that area is called Quadrant 1. Say if 13 was negative instead of going horizontal east you would go horizontal west. And the point should end up in Quadrant 2. Any more questions feel free to ask me below.
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
