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Luden [163]
3 years ago
13

PLEASE HELP ME ASAP PLEASE

Mathematics
1 answer:
Marta_Voda [28]3 years ago
7 0
Answer
Choice D
Company 2 is greater in avg rate of change by $750

=======================================

Explanation:

Average rate of change is the same as slope through two points. The two points are the endpoints of the interval. 

The x values we'll use is x = 0 and x = 12 since it constitutes the first 12 years (12-0 = 12)

For company 1, if x = 0 and x = 12, then y = 12000 and y = 30000 respectively. The two points (0,12000) and (12,30000) are on the graph for company 1
The slope of the line through these two points is
m = (y2-y1)/(x2-x1)
m = (30000-12000)/(12-0)
m = 18000/12
m = 1500
Over the first twelve years, the average rate of change for company 1 is 1500

For company 2, the points (0,15000) and (12,42000) are on the graph
The slope of the line through these two points is
m = (y2-y1)/(x2-x1)
m = (42000-15000)/(12-0)
m = (27000)/(12)
m = 2250
Over the first twelve years, the average rate of change for company 2 is 2250

Now find the difference in the slopes
2250-1500 = 750

Company 2's rate of change is larger by $750
So that's why the answer is choice D. Over this timespan, company 2 is growing faster on average compared to company 1

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2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.
Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by h(t)= -16t^2+15t + 12

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled  h(t)= -16t^2+15t + 12 ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

h'= -32t+15

For critical point put h'=0,

-32t+15=0

t=\frac{15}{32}

t=0.46875

Derivate again w.r.t. t,

h''= -32

It is maximum at t=0.46875.

Substitute t in equation (1),

h(0.46875)= -16(0.46875)^2+15(0.46875)+12

h(0.46875)= -3.515625+7.03125+12

h(0.46875)= 15.515625

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

-16t^2+15t + 12=0

Apply quadratic formula, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Here, a=-16 , b=15, c=12

t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}

t=\frac{-15\pm\sqrt{225+768}}{-32}

t=\frac{-15\pm\sqrt{993}}{-32}

t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}

t=−0.515,1.453

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0
3 years ago
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