The back of Tom's property is a creek. Tom would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a pasture. If there is 920 feet of fencing available, what is the maximum possible area of the pasture

Max Area = 105,800 sq.ft

Step-by-step explanation:

A square will always give us the maximum area.

Thus, one side would be;

920/4 = 230 feet

So, we want a square 230 ft by 230 ft

however, from the question, we are to use the creek as one side. So, we'll take the 230 ft that we don't need because of the creek and then add it to the opposite side to get 230 + 230 = 460 ft.

Thus,we now have a rectangle with dimensions: 230 ft by 460 ft

Area is given by;

area = length × width

Maximum Area = 230 × 460

Max Area = 105,800 sq.ft

529 and 531

step-by-step explanation:

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841