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amm1812
3 years ago
9

How many ways can a person select 3 videotapes from a collection of 10 tapes?

Mathematics
1 answer:
lesya [120]3 years ago
6 0
If the 10 videotapes are distinct (different) and the order of selection is immaterial, then there are
C(10,3)=10!/(3!(10-3)!)=10!/(3!7!)=10*9*8/(1*2*3)=120 
ways to select three videotapes out of 10.
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Which equation represents the slope-intercept form of the line below? y intercept: 0, 8 slope : 1/2
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Answer:

y = 1/2 x + 8

Step-by-step explanation:

We are given the y intercept and the slope, so we can write the equation using the slope intercept form

y = mx+b  where m is the slope and b is the y intercept

y = 1/2 x + 8

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How do i know when a set of ordered pairs that represents a function?
pav-90 [236]

Answer:

How do you figure out if a relation is a function? You could set up the relation as a table of ordered pairs. Then, test to see if each element in the domain is matched with exactly one element in the range. If so, you have a function!

Step-by-step explanation:

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If the ribbon cost $0.10 per foot what is the total cost of the ribbon in dollars
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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
3 years ago
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