jack works after school. each day he earns a set amount, plus an hourly wage. write a linear function f jack can use to determin
e his pay. QUICK
1 answer:
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Part 1) <span>Create a diagram that shows Kamila’s rectangular lawn and the gravel border. Assign variables to any unknown sides and label the diagram
see the attached figure
Part 2)</span><span>Use your diagram to determine how many square feet of the gravel border will surround the lawn
area of the border=[162*92]-[150*80]----> 14904-12000-----> 2904 ft</span>²
Part 3) <span>Kamila decided that the depth of the gravel needs to be 2 inches. What is the total volume of gravel needed for the border?
we know that
1 ft-------> 12 in
x--------> 2 in
x=2/12------> x=0.17 ft
volume of gravel needed=area of the border*deep of the gravel
</span>volume of gravel needed=2904*0.17----> 493.68 ft³
<span>
Part 4)</span><span>A large bag of colored rock contains a cubic yard and costs $30. What is the total cost of the rocks needed for the border?
</span>
1 yd³--------> 27 ft³
x---------> 493.68 ft³
x=493.68/27-----> 18.28 yd³
if 1 yd³----------> cost $30
18.28 yd³------> x
x=18.28*30------> x=$548.53
the answer Part 4) is
$548.53
see the attached figure
Part 2)</span><span>Use your diagram to determine how many square feet of the gravel border will surround the lawn
area of the border=[162*92]-[150*80]----> 14904-12000-----> 2904 ft</span>²
Part 3) <span>Kamila decided that the depth of the gravel needs to be 2 inches. What is the total volume of gravel needed for the border?
we know that
1 ft-------> 12 in
x--------> 2 in
x=2/12------> x=0.17 ft
volume of gravel needed=area of the border*deep of the gravel
</span>volume of gravel needed=2904*0.17----> 493.68 ft³
<span>
Part 4)</span><span>A large bag of colored rock contains a cubic yard and costs $30. What is the total cost of the rocks needed for the border?
</span>
1 yd³--------> 27 ft³
x---------> 493.68 ft³
x=493.68/27-----> 18.28 yd³
if 1 yd³----------> cost $30
18.28 yd³------> x
x=18.28*30------> x=$548.53
the answer Part 4) is
$548.53
Answer:
(a) 720 -2x
(b) 0 ≤ x ≤ 360
(c) A = x(720 -2x)
(d) (30.334, 68.645) ∪ (291.355, 329.666) (two disjoint intervals)
(e) x = 180 ft, the other side = 360 ft; total area 64,800 ft²
Step-by-step explanation:
(a) The two parallel sides of the fenced area are each x feet, so the remaining amount of fence available for the third side is (720 -2x) ft. Then ...
length = 720 -2x
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(b) The two parallel sides cannot be negative, and they cannot exceed half the length of the fence available, so ...
0 ≤ x ≤ 360
__
(c) Area is the product of the length (720-2x) and the width (x). The desired function is ...
A = x(720 -2x)
__
(d) For an area of 20,000 ft², the values of x will be ...
20000 = x(720 -2x)
2x² -720x +20000 = 0
x = (-(-720) ±√((-720)² -4(2)(20000)))/(2(2)) = (720±√358400)/4
x = 180 ±40√14 = {30.334, 329.666} . . . feet
For an area of 40,000 ft², the values of x will be ...
x = 180 ±20√31 ≈ {68.645, 291.355} . . . feet
The values of x producing areas between 20,000 and 40,000 ft² will be values of x in the intervals (30.334, 68.645) or (291.355, 329.666) feet.
__
(e) The vertex of the area function is at the axis of symmetry: x = 180. The corresponding dimensions are ...
180 ft × 360 ft
and the area of that is 64,800 ft².
