Can someone help me please
1 answer:
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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:
Answer: The time taken for the baton to hit the ground is t = 4 mins
Given:
In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.
To find:
The measure of angle P.
Solution:
According to the Law of Cosines:
Using Law of Cosines in triangle OPQ, we get
On further simplification, we get
Therefore, the measure of angle P is 79 degrees.
Notice that every pair of point (x, y) in the original picture, has become (-y, -x) in the transformed figure.
Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.
Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)
so we are very close to (-y, -x).
The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.
ANSWER:
"<span>a 90 clockwise rotation about the origin and a reflection over the y-axis</span>"
Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.
Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)
so we are very close to (-y, -x).
The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.
ANSWER:
"<span>a 90 clockwise rotation about the origin and a reflection over the y-axis</span>"