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dedylja [7]
3 years ago
14

There is a sidewalk of width x around a rectangular garden. If the garden measures twenty-feet by thirty-feet, then the combined

area of the garden and sidewalk is
Mathematics
1 answer:
prohojiy [21]3 years ago
5 0

Answer:

Area = 200 + 50 + x

Step-by-step explanation:

Given

Length = 20

Width = 30

Side Walk = x

Required

Determine the area.

To do this we need to get the new worth and length by adding the length of the sidewalk.

This gives:

Length = 20 + x.

Width = 30 + x

So, area becomes.

Area = (20 + x)(30 + x)

Area = 600 + 20x + 30x + x²

Area = 200 + 50 + x²

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Consider the following.
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Answer:

Step-by-step explanation:

f(t) = t + cos(t)

Criticals :

f' (t) = 1 - sin (t) = 0

sin (t) = 1

Given that the interval is : [ - 2π , 2π ]

thus;  t = π/2 and -3π/2

The region then splits into [ -2π , -3π/2 ], [-3π/2 , 2π ]  and [ π/2 , 2π ]

Region 1:                            Region 2:                         Region 3:

[ -2π , -3π/2 ]                      [-3π/2 , 2π ]                      [ π/2 , 2π ]

Test value (t): -11 π/6          Test value(t) = 0              Test value = π

f'(t) = 1 - sin(t)                       f'(t) = 1 - sin(t)                   f'(t) = 1 - sin(t)

f'(t) = 1 - sin (-11 π/6)            f'(t) = 1 - sin (0)                  f'(t) = 1 - sin(π)

f'(t) =  positive value           f'(t) = 1 - 0                         f'(t) = 1 - 0.0548

thus; it is said to be            f'(t) = 1  (positive)              f'(t) = 0.9452 (positive)

increasing.                          so it is increasing            so it is increasing

So interval of increase is :  [ -2π , 2π ]

There is no  local maximum value or minimum value since the function increases monotonically over [ -2π , 2π ]. Hence, there is no change in the pattern.

c) Inflection Points;

Given that :

f'(t) = 1 - sin (t)

Then f''(t) = - cos (t) = 0

within [ -2π , 2π ], there exists 4 values of  t for which costs = 0

These are:

[-3π/2 ]

[-π/2 ]

[π/2 ]

[3π/2 ]

For Concativity:

This splits the region into [ -2π , -3π/2], [ -3π/2 ,  -π/2], [-π/2 , π/2] , [π/2 , 3π/2] and [ 3π/2 , 2π].

Region 1: [ -2π , -3π/2]      Region 2: [ -3π/2 ,  -π/2]            

Test value = - 11π/6           Test value = π                            

f''(t) = - cos (t)                      f''(t) = - cos (t)

- cos (- 11π/6) = negative    f''(-π) = - cos (- π)

Thus; concave is down.      f''(π) = -cos (π)

                                           positive, thus concave is up

Region (3):   [-π/2 , π/2]

Concave is down

Region (4):  [π/2 , 3π/2]

Concave is up

Region (5):  [ 3π/2 , 2π]

Concave is down

We conclude that:

Concave up are at region 2 and 4:  [ -3π/2 ,  -π/2] ,  [π/2 , 3π/2]

Concave down are at region 1,3 and 5 :  [ -2π , -3π/2] , [-π/2 , π/2] , [ 3π/2 , 2π]

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