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Sauron [17]
2 years ago
5

41. WRITE a Real-Life Problem to fit the data shown in the GRAPH.

Mathematics
1 answer:
galina1969 [7]2 years ago
6 0

The graph is a straight line graph that has a y-intercept of -300 (0, -300),

and an x-intercept of 30, (30, 0).

  • First Part: A Real-Life problem to fit the data is; <u>Elevation of a diving bell</u> from a 300 feet depth at a constant rate.

  • Second Part: The domain of the function is <u>continuous</u>.

Reasons:

First Part;

A Real-life Problem to fit the data shown in the GRAPH is the elevation of a

diving bell from a depth of 300 feet below the surface of the water, at a

constant rate of 10 feet per second.

\begin{array}{|cc|c|} \mathbf{Amount \ owed}&&\mathbf{Time  \ (seconds)}\\-300&&0\\-250&&5\\-200&&10\\-150&&15\\-100&&20\\-50&&25\\0&&30\end{array}

Second Part;

A discrete domain are values of input made up of only particular numbers

contained within an interval, continuous domain are input values that are

made up of all the numbers within an interval.

Given that the graph of the function is a continuous line, and not made up

of a series of dots and that the input value of time can take on any value,

within the domain from 0 to 30 seconds, the domain is a continuous

domain.

Learn more here:

brainly.com/question/25344190

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Cheryl and Daryl are circus performers. A cable lifts Cheryl into the air at a constant speed of 2 ft/s. When Cheryl's arms are
Bas_tet [7]

The statement is incomplete.


This is the full statement, with the question:


Cheryl and Daryl are circus performers. A cable lifts Cheryl into the air at a constant speed of 2 ft/s. When Cheryl's arms are 12 ft above the ground, Daryl throws her a ball. He throws the ball from a height of 6 ft with an initial vertical velocity of 24 ft/s.


Identify Variables


t = time, in seconds, since the ball was thrown

h = height, in feet, above the ground


Cheryl's height = initial height + change in height


h=12+2t


Projectile Motion Formula: −16t2+vt+h0


How to Interpret the Solutions to the System


h = 12+2t


h= − 16t² +24t + 6



Answer:


i) the solution of the system is when both ball and arms are at the same height.


ii) that happens two times, first time at 0.375 s of the ball have been launched, at the height of 12.75 ft, and second time 0.1 s after have been launched at the heigh of 14 ft.



Explanation:



1) Identify the question: you just need to explain how the two equations represent the situation and find the solution.


2) The first equation h = 12 + 2t is the heigth of the arms of Cheryl, where:


i) 12 is the height at which they were at the instant the ball was thrown to her and


ii) 2t is the product of the speed of the cable that pulled her (2 ft/s) by the time.


3) The second equation h= − 16t² +24t + 6 is the height of the ball, where:


i) the first term ( -16 t²) is half the gravitational acceleration (-32 ft/s²) times the time squared,


ii) the second term (24t) is the product of the velocity with which the ball was thrown (initial velocity) times the time, and


iii) the third term (6) is the height from which the ball was thrown.


2) Then, the solution of the system is the moment and heigth at whic the ball thrown vertically upward meet the arms of Cheryl: both ball and arms are at the same height.


3) The solution is found by solving the system:


h = h ⇒ 12+2t = − 16t² +24t + 6


These are the steps:


i) transpose all the terms to one side:


− 16t² + 24t + 6 - 12 - 2t = 0


ii) combine like terms:


− 16t² + 22t - 6 = 0


iii) divide by - 2:


8t² - 11 t + 3 = 0


iii) complete squares and factor the perfect square trinomial:


8t² - 11t = - 3


8 (t² - 11/8 t) = - 3


8(t² - 11/8t + 121/256) = - 3 + 8(121/256)


8(t - 11/16)² = - 3 + 121/32


iv) solve for t:


8(t - 11/16)² = 25/32


(t - 11/16)² = 25/256


t - 11/16 = (+/-) 5/16


t = 11/16 (+/-) 5/16


t = 16/16 = 1 and t = 6/16 = 0.375


Then, the ball meets the arms of Cheryl first time at 0.375 s of have been launched.


And the height was 12 + 2(0.375) = 12.75 ft


The ball meets the arms of Cheryl second time at 0.1 s after have been launched and the heigh was 12 + 2(1) = 14 ft.

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If the 12 is the base, then we have the rectangle with the area of 5(10) = 50 and two triangles each with an area of 1/2(12)(10) = 60:

50+60+60 = 170 sq. units.

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