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3 years ago

Please help me with this problem, and can you also show steps. thank you

Mathematics

2 answers:

Hatshy [7]3 years ago

8 0

((-8 2/5)/(-4 1/10))*(-1/2)
(-(8*5+2)/5)/(-(4*10+1)/10)*(-1/2)
(-(42/5))/(-(41/10))*(-1/2)
((42/5)/(41/10))*(-1/2)
((42/5)*(10/41))*(-1/2)
((42*10)/(5*41))*(-1/2)
((420)/(205))*(-1/2)
(84/41)*(-1/2)
-((84*1)/(41*2))
-((84)/(82))
=-(42/41)
=-42/41
=-1 1/41

Maurinko [17]3 years ago

7 0

Convert all the improper fractions into mixed numbers, then simplify. Then, you do the operations.

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Find the inverse of f (x) = x + 2 + 9

Brut [27]

Answer:

y = x - 11

Step-by-step explanation:

First, turn the f(x) into a y:

y = x + 2 + 9

Now, switch the x and y.

x = y + 2 + 9

Add like terms.

2 + 9 = 11

x = y + 11

Now, solve for y. Subtract 11 on both sides.

y = x - 11

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Step-by-step explanation:

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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

The equation for line r can be written as y=-1/4x+3. Line s which is parallel to lone r includes the point (-4,3). What is the e

MaRussiya [10]

Answer:

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

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when x is 0)
Parallel lines always have the same slope

1) Determine the slope of line S using line R (m)

$y=-\frac{1}{4} x+3$

We can identify clearly that the slope of the line is $-\frac{1}{4}$ , as it is in the place of m. Because parallel lines always have the same slope, the slope of line S would also be $-\frac{1}{4}$ . Plug this into $y=mx+b$ :