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Delicious77 [7]
3 years ago
6

Which comparison is correct?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Option B: $4.5 \times 10^{-6}>3.9 \times 10^{-9}$ is the correct answer.

Explanation:

Option A: $7.1 \times 10^{8}>1.2 \times 10^{9}$

Simplifying, we get,

710000000>1200000000

Since, the value of $1.2 \times 10^{9}$ is greater than $7.1 \times 10^{8}$, the comparison $7.1 \times 10^{8}>1.2 \times 10^{9}$ is false.

Hence, Option A is not the correct answer.

Option B: $4.5 \times 10^{-6}>3.9 \times 10^{-9}$

Simplifying, we get,

0.0000045>0.0000000039

Since, the value of $4.5 \times 10^{-6}$ is greater than $3.9 \times 10^{-9}$, the comparison $4.5 \times 10^{-6}>3.9 \times 10^{-9}$ is true.

Hence, Option B is the correct answer.

Option C: $2.4 \times 10^{-5}

Simplifying, we get,

0.000024

Since, the value of $2.4 \times 10^{-5}$ is greater than $5.7 \times 10^{-6}$, the comparison $2.4 \times 10^{-5} is false.

Hence, Option C is not the correct answer.

Option D: $8.2 \times 10^{-4}

Simplifying, we get,

0.00082

Since, the value of $8.2 \times 10^{-4}$ is greater than $3.6 \times 10^{-6}$, the comparison $8.2 \times 10^{-4} is false.

Hence, Option D is not the correct answer.

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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':

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Zero Product Property:

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Solve for each case.

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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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