Answer:
The sum of this entire line is 180, so to solve this we would subtract 20 from 180, which means x = 160.
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Answer:
the answer is 25 in this question hope it helped
Answer:
4 cm
Step-by-step explanation:
The midpoint segment has length ...
... L = AB/2 + BC + CD/2 = 16 cm
The entire segment has length ...
... AD = AB + BC + CD = 28 cm
If we subtract AD from 2L, we have ...
... 2L - AD - 2L = 2(AB/2 +BC + CD/2) - (AB +BC +CD) = 2(16 cm) -28 cm = 4 cm
... AB +2BC + CD -AB -BC -CD = 4 cm . . . . remove parentheses
... BC = 4 cm . . . . simplify
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.
