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2 years ago

What is the length of line segment DC

Mathematics

2 answers:

Misha Larkins [42]2 years ago

4 0

The answer is 3 units.

12/4= 3

9/3= 3 units

Y_Kistochka [10]2 years ago

3 0

Answer:

The length of DC is 3 unit.

Step-by-step explanation:

Consider the provided figure.

DE is parallel to CB.

According to basic proportionality theorem:

"If a line is drawn parallel to one side of a triangle to pass the other two sides in separate points, the other two sides are divided into the same ratio."

Therefore

$\frac{AE}{EB}=\frac{AD}{DC}$

$\frac{12}{4}=\frac{9}{x}$

$3=\frac{9}{x}$

$x=\frac{9}{3}$

$x=3$

Thus, the length of DC is 3 unit.

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4x – 3y=8<br> 2. 5x – 2y = -11

AlladinOne [14]

Answer:

$x = -7$

$y = - 12$

Step-by-step explanation:

The given system is:

4x-3y=8

5x-2y=-11

We make x the subject in the top equation to get:

$x = \frac{3}{4}y + 2$

Plug this expression for x in the bottom equation to get

$5( \frac{3}{4}y + 2) - 2y= - 11$

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$\frac{15}{4} y + 10 - 2y = - 11$

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Group similar terms

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3 years ago

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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 :)

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2 years ago