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Butoxors [25]
3 years ago
8

A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long

does it take for the ball to land on the ground
Mathematics
1 answer:
wel3 years ago
5 0

Answer: 0.3m/s

Step-by-step explanation:

Height/ velocity

3m/12m per sec

Time= 0.25metres per seconds

To the nearest tenth, 0.3m per seconds

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There are (4^9)5 ⋅ 4^0 books at the library. What is the total number of books at the library?
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7 0
3 years ago
URGENT! PLEASE HELP.
uysha [10]

Answer: a) BC = 1386.8 ft

b) CD = 565.8 ft

Step-by-step explanation:

Looking at the triangle,

AD = BD + 7600

BD = AD - 7600

Considering triangle BCD, we would apply the the tangent trigonometric ratio.

Tan θ = opposite side/adjacent side. Therefore,

Tan 24 = CD/BD = CD/(AD - 700)

0.445 = CD/(AD - 700)

CD = 0.445(AD - 700)

CD = 0.445AD - 311.5 - - - - - - - -1

Considering triangle ADC,

Tan 16 = CD/AD

CD = ADtan16 = 0.287AD

Substituting CD = 0.287AD into equation 1, it becomes

CD = 0.445AD - 311.5

0.287AD = 0.445AD - 311.5

0.445AD - 0.287AD = 311.5

0.158AD = 311.5

AD = 311.5/0.158

AD = 1971.52

CD = 0.287AD = 0.287 × 1971.52

CD = 565.8 ft

To determine BC, we would apply the Sine trigonometric ratio which is expressed as

Sin θ = opposite side/hypotenuse

Sin 24 = CD/BC

BC = CD/Sin24 = 565.8/0.408

BC = 1386.8 ft

4 0
3 years ago
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