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liubo4ka [24]
3 years ago
9

A proportional relationship is shown in the table below: what is the slope and what numbers do i graph. ty besties

Mathematics
1 answer:
dangina [55]3 years ago
5 0

Answer:

The slope of the line is m = \frac{3}{5} and the equation of the line is y = \frac{3}{5}\cdot x.

Step-by-step explanation:

From table we understand that slope of the line is constant, that is, for each change of x there is one and only change in y, fulfilling the main characteristic of a line and we can determine the slope for every point of the line by means of the following expression:

m = \frac{y_{f}-y_{o}}{x_{f}-x_{o}} (1)

Where:

x_{o}, x_{f} - Initial and final x-coordinates of the line.

y_{o}, y_{f} - Initial and final y-coordinates of the line.

If we know that (x_{o}, y_{o}) = (0,0) and (x_{f},y_{f}) = (8,4.8), then the slope of the line is:

m = \frac{4.8-0}{8-0}

m = \frac{3}{5}

From graph we conclude that x-intercept of the line is b = 0. Hence, the equation of the line is y = \frac{3}{5}\cdot x and we proceed to plot the expression with the help of a graphing tool.

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3 years ago
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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Answer:

180

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Answer:

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