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Zarrin [17]
3 years ago
5

A dart is thrown at a dartboard like the one below. If the dart

Mathematics
1 answer:
kolezko [41]3 years ago
3 0

Answer:

<h2>brainiest me</h2>

Step-by-step explanation:

a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square? Solution. If the side lengths of the dart board and the side lengths of the center square are all then the side length of the legs of the triangles are . Use Geometric probability by...

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asambeis [7]
The answer is 35 hope this helps
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3 years ago
I'm don't understand polygons, as you should know
lawyer [7]

Answer:

540°

Step-by-step explanation:

The sum of the interior angles of a polygon is

sum = 180° (n - 2) ← n is the number of sides

Here n = 5, thus

sum = 180° × 3 = 540°

8 0
3 years ago
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Answer:

1. tan (4π/3) = √3

Answered by GAUTHMATH

5 0
3 years ago
Lin uses an app to graph the charge on her phone.
exis [7]

Answer:

  • See below

Step-by-step explanation:

<u>Use the graph to answer the following questions:</u>

When did she start using her phone?

  • At 2 afternoon

When did she start charging her phone?

  • At 8 afternoon

While she was using her phone, at what rate was Lin’s phone battery dying?

<u>From 100% to 40% between 2PM and 4 PM:</u>

  • (100 - 40)/(4 - 2) = 60/2 = 30% per hour
8 0
2 years ago
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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 :)

5 0
3 years ago
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