1. A new factory is being built in our Region on a lot that measures 50 metres by 30 metres. The nearest town's by-law states that a
lawn of uniform width must surround the factory such that the
area of the lawn is equal to the area of the factory. If the area of
the factory is 750 m² what must the width of the lawn be correct to
one decimal place?

21000

Step-by-step explanation:

Let x = the width of the lawn

Find the total area of the factory and the lawn: 200 * 210 = 42000 sq/ft

Factory area has to equal lawn area, therefore

21000 sq/ft = area of the lawn and the factory

The dimensions of the factory: (200-2x) by (210-2x)

Find x using the area of the factory

(200-2x) * ( 210-2x) = 21000

42000 - 420x - 400x + 4 = 21000

<=>  - 205x + 5250 = 0

<=> (x-30)(x-175) = 0

<=> x= 30 or x=175 but x = 30 ft is the solution that makes sense for the width of the lawn, so

Length: 210 - 2(30) = 150 ft

Width: 200 - 2(30) = 140 ft

Thus we have the area of the factory using these dimensions:

150 * 140 = 21000

Step-by-step explanation:

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40