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statuscvo [17]
3 years ago
14

HURRY I'LL PICK A BRAINLIEST!!!

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
6 0
K = 4a+9ab
k = a(4+9b)
k/(4+9b) = a
a = k/(4+9b)

None of the answer choices match what I got, so I'm thinking there's a typo. Perhaps choice A is supposed to say a = k/(4+9b) since that seems like the closest
Ber [7]3 years ago
6 0

Answer:

Step-by-step explanation:

A

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1b. What is g(-5)? *
ivolga24 [154]

Answer:

-5g

Step-by-step explanation:

5 0
3 years ago
The numerator and the denominator of a certain fraction are consecutive odd numbers. If nine is subtracted frm the numerator, th
Pachacha [2.7K]

The denominator of a fraction is 1 more than 3 times the numerator. If the denominator is doubled and the numerator is increased by 2, the value of the resulting fraction is 1/4. Find the original fraction.

4 0
4 years ago
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What value of s makes the following equation true?<br><br> s3 = −343
vovikov84 [41]
-144.33333 or -14 1/3
3 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
Let f(x)= 12/4x+2 find f(-1)
denis23 [38]
F(x) = 12/(4x+2)

x = -1

f(-1) = 12 / [4(-1)+2]

f(-1) = 12 / (-4+2)

f(-1) = 12 / -2

f(-1) = -6
5 0
3 years ago
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