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vladimir1956 [14]
3 years ago
14

(x^-4)^5 please simplify the expression

Mathematics
1 answer:
pogonyaev3 years ago
4 0
(x ^{-4} )^{5}

Apply exponent rule :

(a^b)^c = a^ ^{b.c}

=x ^{(-4)*5}

(-4)*5

= - 20

Apply exponent  rule :

a ^{-b} =  \frac{1}{a^b}

=  \frac{1}{x ^{20} }

hope this helps!
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3x-9-2x+4. answer????
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The answer is x-5 First collect like term so it will be x-9+4 and then subtract 9-4= 5 so now it will be x-5
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$1111 at 11% for 11 years​
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Answer:

$1344.31

Step-by-step explanation:

1111*0.11*11=1344.31

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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
Each year a certain type of tree grows vertically 14/15 as much as it did the year before. If the tree grows 1.5m during the fir
Anastaziya [24]

https://answers.yahoo.com/question/index?qid=20180411132938AApZYZz

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3 years ago
50 POINTS PLEASE HELP
Rainbow [258]

first simplify the numerator (x+3)(x-2)=x^2-2x+3x-6=x^2+x-6

now factor the numerator (x+3)(x-2)

now find the restriction 3x-6=0  3x=6  x=2

now factor the denominator 3(x-2)

((x+3)(x-2))/3(x-2)

now cross off the (x-2)

and you are left with (x+3)/3

ANSWERS:

The restriction is X=2

The simplified fraction is (x+3)/3

7 0
3 years ago
Read 2 more answers
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