Mathematics, 12.11.2019 23:31, kuddlebugsmommy

Need on this question

the linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds:

a linear model with ordered pairs at 0, 60 and 2, 80 and 3, 80 and 4, 20 and 6, 0 and 7, 0 and 8, 0. the x axis is labeled time in seconds, and the y axis is labeled height in feet.

part a: during what interval(s) of the domain is the water balloon's height increasing?

part b: during what interval(s) of the domain is the water balloon's height staying the same?

part c: during what interval(s) of the domain is the water balloon's height decreasing the fastest? use complete sentences to support your answer.

part d: use the constraints of the real-world situation to predict the height of the water balloon at 10 seconds. use complete sentences to support your answer.

Answers: 2

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Answers

answered: fangirl2837

Part A) Interval [0,2]

Part B) Interval [2,3] and [6,8]

Part C) In the interval [3,4] the water balloon's height decreasing the fastest

Part D) The water balloon's height at 10 seconds is equal to zero

Step-by-step explanation:

Part A: During what interval(s) of the domain is the water balloon's height increasing?

see the attached figure to better understand the problem

During the interval [0,2] the height of the water balloon is increasing

All real numbers greater than or equal to 0 seconds and less than or equal to 2 seconds

The slope of the linear equation in that interval is positive

Part B: During what interval(s) of the domain is the water balloon's height staying the same?

Looking at the graph

During the interval [2,3] and [6,8] the water balloon's height is the same

The slope of the linear equation in that intervals is equal to zero ( is a horizontal line)

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest?

Determine the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Interval [3,4]

For x= 3 sec ----> h=80 ft

For x= 4 sec ----> h=20 ft

substitute

$m=\frac{20-80}{4-3}\\m=-60\ ft/sec$

Interval [4,6]

For x= 4 sec ----> h=20 ft

For x= 6 sec ----> h=0 ft

substitute

$m=\frac{0-20}{6-4}\\m=\frac{-20}{2}\\m=-10\ ft/sec$

therefore

In the interval [3,4] the water balloon's height decreasing the fastest, because the graph has the steepest negative slope

Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 10 seconds

Looking at the graph

During the interval [6,8]

The water balloon's height at 10 seconds is equal to zero, because the water balloon hit the ground at x=6 sec

Need on this question the linear model represents the height, f(x), of a water balloon thrown off

answered: Guest

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function $y=f(x)$ the average rate of change from point $a(a,f(a))$ to point $b(b,f(b))$ can be calculated as

$\text{average rate of change}=\dfrac{f(b)-f(a)}{b-a}.$

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,

$\text{average rate of change}=\dfrac{2-8}{10-2}=\dfrac{-6}{8}=-\dfrac{3}{4}.$