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iVinArrow [24]
2 years ago
5

The figure represents a traffic island that has angles measuring 60°, 20°, and 100°. Enter each angle next to its correct measur

e.
PLZZ HELPPPP
Mathematics
2 answers:
Radda [10]2 years ago
7 0

Answer:

m<M =100 M<N=60 m<p=20

Step-by-step explanation:

suter [353]2 years ago
3 0

Answer:

possblie awnser is 1:Possible answer for Triangle 1: m∠A = 70°; m∠B = ∠55°; m∠C = 55°.

The compass marks equal lengths on both sides of ∠A; therefore, AB

Step-by-step explanation:

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Find the x and y intercept of each line y=7x + 5
Zina [86]

Answer:

y intercept is 5. x intercept is 5/7

4 0
3 years ago
A boat that sold for $12,500 had a sales tax of $562.50. How much is the tax on a boat that sells for $9,550?
Degger [83]

Answer:

$429.75

Step-by-step explanation:

First, find the tax rate on the sale of the boat:  Let r represent that rate.  Then,

$12,500r = $562.50.  Solving for r, we get r = $562.50 / $12,500 = 0.045

The tax rate is 0.045, or 4.5%.

Applying this tax rate to a boat selling for $9,550:

0.045($9,550) = $429.75.  This is the amount of tax on the 2nd boat.

7 0
2 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
2 years ago
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