Farmer Ed has 9,500 meters of fencing, and wants to enclose a rectangular plot
that borders on a river. If Farmer Ed
does not fence the side along the river,
what is the largest area that can be
enclosed?
х
9,500 –2x

The largest area that can be enclosed is
square meters.
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Because it is a rectangle, the area is expressed as A = xy, or length times width.

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get:

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get:

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for x

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for xx = 7500 - 2y substitute into the area equation

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for xx = 7500 - 2y substitute into the area equationA = (7500 - 2y)y distribute

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for xx = 7500 - 2y substitute into the area equationA = (7500 - 2y)y distributeA = -2y2 +7500y

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for xx = 7500 - 2y substitute into the area equationA = (7500 - 2y)y distributeA = -2y2 +7500y

Because it is a rectangle, the area is expressed as A = xy, or length times width.Because it is next to the river, he only needs to fence three sides, so F = x + 2y.Knowing the amount of fencing available is 7500m, we get: 7500 = x + 2y solve for xx = 7500 - 2y substitute into the area equationA = (7500 - 2y)y distributeA = -2y2 +7500y You can see that this is a parabola which opens down, meaning that the point of maximum area will be at the vertex, y = - b/2a = -7500/[2(-2)] = 1875

/2a = -7500/[2(-2)] = 1875

/2a = -7500/[2(-2)] = 1875 x = 7500 - 2(1875) = 3750

/2a = -7500/[2(-2)] = 1875 x = 7500 - 2(1875) = 3750

/2a = -7500/[2(-2)] = 1875 x = 7500 - 2(1875) = 3750 A = 3750(1875) = 7,031,250 m2

/2a = -7500/[2(-2)] = 1875 x = 7500 - 2(1875) = 3750 A = 3750(1875) = 7,031,250 m2

3.4 pizzas

step-by-step explanation:

given is a scatter plot showing number of pizzas sold during weeks for different issues of coupon

y represents number of pizzas sold and x number of coupons issued

y is marked verticaly and x horizontally

the trend line is given as

y =3.4x+43

this says that even when no coupon is issued pizzas sold will be 43

and rate of change of pizza sold for 1 unit of coupon = slope = 3.4

i.e. when x increases by 1 unit y increases by 3.4 units

hence for each additional coupon issued 3.4 pizzas more will be sold

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3