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10.03.2020 •
Mathematics
In the air, it had an average speed of 16m/s . In the water, it had an average speed of 3 m/s before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone's entire fall took 12 seconds.How long did the stone fall in air and how long did it fall in the water?
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Ответ:
Time in the air t₁ = 7 s
Time in the sea t₂ = 5 s
Step-by-step explanation:
We will use the equation: v = d/t , where:
v is the average speed,
d the distance traveled, and
t time doing the trip
From v = d/t ⇒ d = v*t
Therefore, if we call t₁ time in the air, and t₂ time in the sea, according to problem statement, we have:
t₁ + t₂ = 12 (1)
16*t₁ + 3*t₂ = 127 (2)
A two equation system, from equation (1) we get
t₂ = 12 - t₁
And by substitution in equaton (2)
16*t₁ + 3 * ( 12 - t₁ ) = 127
16*t₁ + 36 - 3*t₁ = 127
13*t₁ = 91
t₁ = 91 /13
t₁ = 7 s
And t₂ = 12 - 7
t₂ = 5 s
Ответ:
2) C. (LINEAR PAIR THEOREM)
3) (D. TRANSITIVE PROPERTY OF CONGRUENCY) to equations (1) and (2), we get m∠IKL + m∠JLD = 180°.
4) C. (CONGRUENT SUPPLEMENTS THEOREM