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nexus9112 [7]
3 years ago
8

Officials project that between 2010 and 2050, the population of a certain city will drastically change. The model below gives th

e projection of the city's population , P, in thousands, with respect to time, t, in years, where 2010 corresponds to t = 0.P=175 + (11/2)tWhat does the 175 mean in the equationin 2010 the population of the city was 175 thousand
in 2050 the population of the city will be 175 thousand
Between 2010 and 2050, the population of the city will increase by 175 thousand
Between 2010 and 2050 the population of the city will decrease by 175 thousand
Mathematics
1 answer:
MaRussiya [10]3 years ago
3 0

Answer:

In the year 2010, the population of the city was 175,000

Step-by-step explanation:

If we rewrote this as a linear expression in standard form (it is linear, btw), it would look like this:

P(t)=\frac{11}{2}t+175

The rate of change, the slope of this line, is 11/2.  If the year 2010 is our time zero (in other words, we start the clock at that year), then 0 time has gone by in the year 2010.  In the year 2011, t = 1 (one year goes by from 2010 to 2011); in the year 2012, t = 2 (two years have gone by from 2010 to 2012), etc.  If we plug in a 0 for t we get that y = 175,000.  That is our y-intercept, which also serves to give us the starting amount of something time-related when NO time has gone by.

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CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is
Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

\cos(x)=-\sin(x)

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

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