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xeze [42]
4 years ago
10

I PUIUL

Mathematics
1 answer:
MrRa [10]4 years ago
8 0

george weighs eight more pounds than igor.

14-6=8 would be the equation to use

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Choose the lowest common denominator.
Brilliant_brown [7]

1) 8

2) -\frac{3}{4}=-\frac{6}{8}, so it is equal to -\frac{6}{8}+\frac{5}{8}

3) 1/8

4 0
2 years ago
Mai-Li has 5 1/8 yd of material her new shirt will take 2 3/4 yd how much material will she have left after the skirt is made.
AysviL [449]
The correct answer is C. 2 3/8 yd.


I hope that helped you! c:
5 0
4 years ago
Read 2 more answers
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
Which of the following integrals cannot be evaluated using simple substitution?
adell [148]

the second and the last,

since quadratic inside radical without multiplication with linear function

8 0
3 years ago
ANYONE PLEASE HELP ME WITH MY MATH HOMEWORK I REALLY NEED THE ANSWER RIGHT NOW BECAUSE I HAVE TO PASS THIS LATER I DON’T HAVE MU
devlian [24]

Answer:

19.) \frac{y^3}{2x^{2} }

20.) \frac{3s}{7t^3}

21.) 23a^9b^8

22.) 34x^8y^{14}

Step-by-step explanation:

Glad to help

--Chris Jackson [High School Junior: Calculus Senior Honor Student]

5 0
3 years ago
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