2. a farmer is planning to create a temporary ‘pen’ for livestock, by cutting a large section of plastic netting into strips, joining the strips together and then fixing the combined length to posts. he decides the fence must be 2 m high, and must form a shape that is nearly square: either a precise square or a rectangle whose length and width do not differ by more than 5 m. if the length and width are both whole numbers of metres, what is the greatest area than can be created from 156 m2 of netting? (provide computation and logical reasoning, )

Step-by-step explanation:

So this one is a little tricky.  We need to pay attention to the area of the pen, and the strips.  Well, what do we know

area of a rectangle or square is l*w

a square has four equal sides

a rectangle has  two sets of equal sides opposite from each other (top and bottom will be equal and left and right will be equal)

the length and width of the pen cannot differ more than 5 meters

The strips will have an area too, where their height is 2

the sum of the area of the strips will need to be no greater than 156

If you don't understand how I got any of those let me know.

There will be 4 strips total, two will be "length strips and two will be "width strips" basically being along the length  or width of the pen.  However long they are (lets call length strips L and width strips w) they will have a height of 2, so since they will add up to be at most 156 we can write out an addition of the areas

L*2 + L*2 + w*2 + w*2 = 156 we have two of each since there are two lengths and two widths.  We can make this a little simpler though.

4L + 4w = 156 We can further divide both sides by 4

L + w = 39, so now we have a bit more useful information.

Now, the area of the pen is similarly L*w and we want to make this as large as it can be.  It might be more useful though if we only have one variable.

since we know L + w = 39, we also know 39 - w = L, so we can put that w into the area of the pen.  So L*w becomes (39-w)*w or 39w - w^2

Now we have an equation we want to maximize, but! we also want to make sure w is not 5 more or less than l.

This is a quadratic, so there are a few ways  to find the max, I will assume you can, but if not let me know.  

The very maximum is at w = 19.5  It does say that it needs to be a whole number though, so you can pick 19 or 20  What does this make length?  Well, since L = 39-w then L will either be 19 or 20, depending on the number you choose.  Either way, one will be 19 and one will be 20.  And of course you should check if they are within 5 of each other and of course they are.  

Now for the main question, what is the greatest area? well we know we get it with 19*20 and that is 380, so there is your answer.  If there was anything you didn't understand feel free to ask.  

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.