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Law Incorporation [45]
2 years ago
10

The volume of a recta

Mathematics
2 answers:
kumpel [21]2 years ago
5 0
There is no graph to find the volume
Usimov [2.4K]2 years ago
4 0
I can’t see no rectangle
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Mekhanik [1.2K]
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8 0
3 years ago
Find dy/dx by implicit differentiation for ycos(x)=xcos(y)
alekssr [168]

Answer:

dy/dx = (cos y + y sin x) / (cos x + x sin y)

Step-by-step explanation:

y cos x = x cos y

y (-sin x) + dy/dx cos x = x (-sin y dy/dx) + cos y

-y sin x + dy/dx cos x = -x sin y dy/dx + cos y

dy/dx (cos x + x sin y) = cos y + y sin x

dy/dx = (cos y + y sin x) / (cos x + x sin y)

5 0
3 years ago
10TH GRADE GEOMETRY, REAL ANSWERS PLS. HELP :)
lisov135 [29]
(B) is the right answer
5 0
3 years ago
Read 2 more answers
The volume of the rectangular prism is 240 cubic centimeters. A rectangular pyramid has the same length, width,and height as a p
LuckyWell [14K]
The volume of the pyramid would be 80 cubic centimeters. The reason for this is because the formula for volume of a pyramid is 1/3bh, so if all the dimensions are the same for both figures, just divide the volume of the prism by three to find the volume of the pyramid. If you do this 240÷3=80 or the volume of the pyramid.
4 0
3 years ago
Read 2 more answers
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 :)

5 0
2 years ago
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