This backy
dimensions
a circular
un in the
Croydon has a circular pool in his
as shown below 14
20 m
14 m
Pool
10m
15 m-
20 m
How much should Denton be paying
per month
A 550
B $60
6 570
D$80
Lawn
sed to
kol board
14 m
Which of the following is closest to the are
of the lawn suurounding the pool?

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7

  b:   ∠x = 28.6°, ∠y = 31.2°, ∠z = 120.2°

step-by-step explanation:

this is one of those questions where "test taking skill" is all you need. the only answer that has angles that sum to 180° is the one shown above. the remaining choices do not describe the angles of a triangle.

if you want to do more work than simply adding the offered numbers, you can also check them against the law of sines. it should be true that

  x/sin(x) = y/sin(y) = z/sin(z)

and all of these ratios must be greater than 15, the longest side length.

__

if you want to solve the triangle from scratch, i'd suggest solving for the largest angle using the law of cosines.

  z² = x² +y² -2xy·cos(z)

  z = arccos((x² +y² -z²)/(2xy)) = arccos(-75.11/149.4) = 120.182°

now you can find y from the law of sines

  y = arcsin(y/z·sin(z)) = 31.242°