Roy wants to make a path from one corner of his yard to the other as shown below. The path will be 4 feet wide. He wants to find the area of lawn that remains. A rectangle has a height of 40 feet and a base of 15 feet. A 4 feet wide path forms a parallelogram inside of the rectangle. Roy claims that the area of the lawn is 300 square feet since it covers exactly one-half of the yard. Which statement about his claim is correct?

Options

(A)He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 440 sq ft.

(B)He is incorrect. The path will have an area of 1/2(4)(40)=80 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 220 sq ft.

(C)He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 140 sq ft.

(D)He is incorrect. The path will have an area of (9)(40)=360 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 240 sq ft.

(A)He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 440 sq ft.

Step-by-step explanation:

The diagram is attached below.

From the diagram,

The path of width 4 feet divides the yard into two triangles of base 11 feet and height 40 feet respectively. (See the second diagram).

Therefore:

Area of the lawn = 2(Area of One Triangle)

=2(0.5 X 11 X 40)

Therefore, Roy is Incorrect.

The correct option is A.

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!