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Wittaler [7]
3 years ago
9

I can’t figure out what math to do for problem #2

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
4 0
First divide 6/20.
that gives you 3/10.
then multiply by five on the numerator and denominator.

that gets you 15/50.
so, he wins 15 games out of 50.
I hope this helped! :))
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What are the intercepts to the line
Roman55 [17]
The x-intercept is 8 because it passes through the horizontal line (x) at point 8.
The y-intercept is -4 because it passes through the vertical line a (y) at point -4.
The correct answer is B.
Hope this helps!
5 0
3 years ago
This table shows information about two occupations.
ycow [4]

yes it can be possible to do

5 0
3 years ago
Mari used a thermometer to record temperatures of −3.4° Celsius and 1.6° Celsius. Which temperature in degrees Celsius is less t
Studentka2010 [4]

Answer: Qualquer temperatura menor que -3,4°C. Ou seja, -3,5 ; -3,6 ; ...

Step-by-step explanation:

8 0
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 :)

5 0
3 years ago
Pleas help please !!!!!!!!!!!!!!!!!!!!!!!
Zepler [3.9K]

Answer:the first one is 6 and the seconndone is 4.5  

Step-by-step explanation:

x- gose forward 2

y- gose 2.5

8 0
3 years ago
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