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3 years ago

Which points satisfy the following system of linear inequalities? Show all your work. -5y+4x<18 -15y-9x>117

1 answer:

3 0

The lines intersect on the graph at (-3, -6). The solution set falls within the domain of negative infinity to positive infinity and a range of y is greater than or equal to -6. Without having the capabilities of graphing it to show you, that's the best I can do. Graph the lines (dotted, I presume) and see if you can follow those guidelines for the solution set.

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Which number can be added to a rational number to explain that the sum of two rational numbers is rational? A) π B) 3 7/8 C) 8 D

Vikki [24]

I believe the answer would be C

6 0

3 years ago

Read 2 more answers

Helpp please show work!

atroni [7]

Answer:

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7 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

Enter the missing values in the area model to find –3(5.5n - 1) 5.5n -1 -3 According to the model above, -3(5.5n - 1)

Elena-2011 [213]

Answer:

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Step-by-step explanation:

3 0

3 years ago

I think the answer is c can someone check me

Katen [24]

Answer:

Step-by-step explanation:

you first have to find the domain by finding where the expression is defined

so the answer would be x>5

6 0

3 years ago