Scsb17 what is -5 + 8x

Please help me with the correct answer <br><br> Solve (x - 3)^2=5

Rus_ich [418]

The answer is B. First solve the parentheses. Then make a quadratic equation and solve from there.

4 0

3 years ago

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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

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

11 times what give you 165

denis23 [38]

165 divided by 11 equals 15

8 0

3 years ago

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Unit 3 homework 6 Gina Wilson

Sonja [21]

Answer:

5) The equation of the straight line is 2 x - y + 1 =0

6) The equation of the straight line is x + y -5 =0

7) The equation of the straight line is 5 x + 6 y - 24 =0

8) The equation of the straight line is x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

5)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (1,3) , ( -3,-5)

$m = \frac{-5-3 }{-3-1 } = \frac{-8}{-4} = 2$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 3 = 2 ( x - 1 )$

y = 2x - 2 +3

2 x - y + 1 =0

The equation of the straight line is 2 x - y + 1 =0

6)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (1,4) , ( 6,-1)

$m = \frac{-1-(4) }{6-1 } = \frac{-5}{5} = -1$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 1 = -1 ( x - 4 )$

y - 1 = - x +4

The equation of the straight line is x + y -5 =0

7)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (-12 , 14) , ( 6,-1)

$m = \frac{-1-(14) }{6+12 } = \frac{-15}{18} = \frac{-5}{6}$

$y - 14 = \frac{-5}{6} ( x - (-12) )$

6( y - 14 ) = - 5 ( x +12 )

6 y - 84 = - 5x -60

5 x + 6 y -84 + 60 =0

5 x + 6 y - 24 =0

8)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (-4 , -2) , ( 4 , 0)

$m = \frac{0+2}{4 +4} = \frac{2}{8} = \frac{1}{4}$

$y - (-2) =\frac{1}{4} ( x - (-4) )$

4 ( y + 2) = x + 4

x - 4 y -4 =0

9)

The equation of the line y = 3x + 6 is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒ 3x + y + k =0

⇒ 12 + 7 + k =0

⇒ k = -19

The equation of the parallel line is 3x + y -19 =0

4 0

3 years ago

TROOOOOOOOOOOOOOOOOOOOOOOOOOOOOOL GANG

TiliK225 [7]

Let's gooooooooooooiio!

6 0

3 years ago

Scsb17 what is -5 + 8x - 10