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svlad2 [7]
3 years ago
7

Julie and Eric row their boat (at a constant speed) 18 miles downstream for 3 hours helped by the current. Rowing at the same ra

te, the trip back against the current takes 6 hours. Find the rate of the boat in still water
Mathematics
1 answer:
svp [43]3 years ago
5 0

Answer:

answer: the crews rowing rate is 6 mph and the rate of the current is 3 mph

Step-by-step explanation:

v the crew rowing rate

u rate of the current

downstream  v + u = 13.5/1.5 = 9 mph

upstream      v - u = 13.5/4.5 = 3 mph

u = 9 - v

v - ( 9 - v ) = 3

v - 9 + v = 3

2 v = 12

v 6 mph,   u 9 - 6 = 3 mph

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5. If three times a number is increased by 4, the result is -8.
KonstantinChe [14]
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Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives
lapo4ka [179]

Answer:

a) 40 feet

b) 54 ft/min

c) 4 mins

Step-by-step explanation:

Solution:-

- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.

- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:

                           h ( t ) = −16t² + 54t + 40

Which gives the height of the baton from ground at time t mins.

- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.

- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:

                         h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40

                         h ( 0 ) = hi = 0 + 0 + 40 = 40 feet

Answer: The height of the platform hi is 40 feet.

- The speed ( v ) during the parabolic path of the baton also varies with time t.

- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.

                        v ( t ) = dh / dt

                        v ( t )= d ( −16t² + 54t + 40 ) / dt

                        v ( t )= -2*(16)*t + 54

                        v ( t )= -32t + 54

- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.

Hence,

                        v ( 0 ) = vi = -32*( 0 ) + 54

                        v ( 0 ) = vi = 54 ft / min

Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min

- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.

- The height of the ball at ground is zero. Hence,

                     h ( t ) = 0

                     0 = −16t² + 54t + 40

                     0 = -8t^2 + 27t + 20

- Use the quadratic formula to solve the quadratic equation:

                     

                    t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t =  \frac{27 + 37}{16} \\\\t = 4

Answer: The time taken for the baton to hit the ground is t = 4 mins

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The four highest scores on the floor exercise at a gymnastics meet were 9.685, 9.28, 9.366, and 9.5 points.
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Step-by-step explanation:

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I am a little rusty on this but i believe it is C
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