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Lelu [443]
4 years ago
11

A hot-air balloon descends from a height of 2000 feet for 40 seconds. The function models the height of the hot-air balloon duri

ng this descent, where t is the amount of time since the descent began. What is the practical domain of the function in this situation? What is the practical range of the function in this situation?
Mathematics
1 answer:
Zielflug [23.3K]4 years ago
8 0

Answer:

The domain is all real values greater than or equal to 0 seconds and less than or equal to 40 seconds

The range is all real values greater than or equal to 0 feet and less than or equal to 2,000 feet

Step-by-step explanation:

Let

t ---->  is the amount of time since the descent began

f(t) ---> is the height of a hot-air balloon descends

we know that

The linear equation in slope intercept form is equal to

y=mx+b

where

m is the unit rate or slope of the linear equation

b is the y-intercept or initial value

In this problem we have that

The slope or unit rate is equal to

m=-\frac{2,000}{40}=-50\ feet\ per\ second ---> is negative because is a decreasing function

The y-intercept or initial value is equal to

b=2,000\ ft

substitute

f(t)=-50t+2,000

The domain is the interval -----> [0,40]

0\leq t\leq 40

All real values greater than or equal to 0 seconds and less than or equal to 40 seconds

The range is the interval ----> [0,2,000]

0\leq f(t)\leq 2,000

All real values greater than or equal to 0 feet and less than or equal to 2,000 feet

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Answer:

a) cos(\alpha)=-\frac{3}{5}\\

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Step-by-step explanation:

a) The problem tells us that angle \alpha is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}

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sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\

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sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}

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Since sin(\alpha)=\frac{4}{5}

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artcher [175]
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Answer A.

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