Sina - (cosa)(tanb)/cosa + (sina)(tanb)
sina ≡ (tana)(cosa)
(tana)(cosa) - (cosa)(tanb)/cosa + (tana)(cosa)(tanb)
= cosa(tana - tanb)/cosa(1 + tanatanb)
(cosas cancel out)
= (tana - tanb)/(1 + tanatanb) ≡ tan(a-b)
The height of the prism is 12 feet.
Write and solve a compound inequality to model the possible length of the dog run.
The inequality to model the possible length of the dog run is;. 100 ≤ 2.50x ≥ 200
And the possible length of the dog run is 80ft.
Minimum spending = $100
Maximum spending = $200
Cost per square feet = $2.50
let
x = possible number of square feet
The inequality:
100 ≤ 2.50x ≥ 200
This means possible number of square feet constructed is greater than or equal to $100 or less than or equal to $200
solve:
100 ≤ 2.50x ≥ 200
divide the inequality into 2
100 ≤ 2.50x
x ≤ 100/2.5
x ≤ 40
the other part:
2.50x ≥ 200
x ≥ 200/2.50
x ≥ 80
Therefore,
the possible length of the dog run is 80 feet
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Answer:
Given : The high school stadium contains a track that surrounds a soccer field. The soccer field is 100 yards long and 70 yards wide.
The school decided to cover the semicircles with grass to create a space for stretching and other activities.
To Find: area of one of the semicircles at either end of the track
How many square yards of grass will the school need to cover the entire circle?
Solution:
Diameter of semicircle = 70 yards
Radius of semicircle = 70/2 = 35 yards
Area of semicircle at one end = (1/2)πr²
= (1/2)(22/7)35²
= 1,925 sq yards
area of one of the semicircles at either end of the track = 1,925 sq yards
square yards of grass will the school need to cover the entire circle
= 2 x 1,925
= 3850 sq yards
distance around one of the semicircles at either end of the soccer field = πr = (22/7) 35 = 110 yards
distance around the inner lane of the track = 100 + 100 + 110 + 110
= 420 yards
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