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3 years ago

Plz help question below

Mathematics

1 answer:

bearhunter [10]3 years ago

5 0

Plan A costs a total of $95 since it says $95 for unlimited talk and text.

Plan B:
(.10 x 800) + (.05 x 1000)
The (.10 x 800) represents 10 cents per talk minute for 800 minutes.
The (.05 x 1000) represents 5 cents per text message for 1000 text messages.
Solve:
.10 x 800 = 80
.05 x 1000 = 50
80 + 50 = 130
This means Plan B will cost him $130 under these conditions.

Plan C:
20 + ((.05 x 800)+(.05 x 1000))
The 20 + represents a flat rate of $20 per month.
The (.05 x 800) represents 5 cents per call minute.
The (.05 x 1000) represents 5 cents per text.
Solve:
.05 x 800 = 40
.05 x 1000 = 50
20 + 40 + 50 = 110
This means Plan C will cost him $110 under these conditions.

Plan D:
45 + (.10(800 - 500))
The 45 + represents a flat monthly rate of $45.
The (800 - 500) represents how many minutes he has to pay for with the 500 free.
The .10 is the cost per extra minute.
Solve:
800 - 500 = 300
.10 x 300 = 30
45 + 30 = 75
This means Plan D will cost him $75 under these conditions.

In short:
Plan A- $95
Plan B- $130
Plan C- $110
Plan D- $75
The least expensive among these is Plan D, which only costs $75 per month.

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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

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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 :)

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