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erastovalidia [21]
3 years ago
14

What is the volume of a cylinder, in cubic feet, with a height of 7 feet and a base

Mathematics
1 answer:
ruslelena [56]3 years ago
6 0

Answer:

1781.29

Sorry, no explanations

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Hi! can you please help me? please show your answer!
Sergio039 [100]

Answer:

1)  f=−8/7 2) z=−3/7

Step-by-step explanation:

1)-f+8=-8f

−f+8=−8f

Step 1: Add 8f to both sides.

−f+8+8f=−8f+8f

7f+8=0

Step 2: Subtract 8 from both sides.

7f+8−8=0−8

7f=−8

Step 3: Divide both sides by 7.

7f/7=−8/7

f=-8/7

2) 6z-5=-z-8

6z−5=−z−8

Step 1: Add z to both sides.

6z−5+z=−z−8+z

7z−5=−8

Step 2: Add 5 to both sides.

7z−5+5=−8+5

7z=−3

Step 3: Divide both sides by 7.

7z/7=−3/7

3 0
3 years ago
4(5t - 7)=10t+2 how to simplify equation
prohojiy [21]
20t-28=10t+2⇔20t-10t=2+28⇔10t=30⇔t=3
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4 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 :)

5 0
3 years ago
What is 7/8 in decimal
Pachacha [2.7K]
To get a decimal form of a fraction, divide the numerator by the denominator.
7/8 = 0.875
6 0
3 years ago
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