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Alika [10]
3 years ago
9

Which expression is equivalent to

Mathematics
2 answers:
sergejj [24]3 years ago
7 0

Answer:

x^{\frac{1}{8} } y^{8}

Step-by-step explanation:

x^{1/4 \times 1/2} y^{16 \times 1/2}

x^{1/8} y^{8}

aleksley [76]3 years ago
7 0

Answer:

x^1/8 y^8

Please see the attached picture for full solution

Hope it helps

Good luck on your assignment

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5. What is the vertex for y = 3(x+19)^2 - 14?
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Answer:

The vertex is (-19,-14)

Step-by-step explanation:

y = 3(x+19)^2 - 14

The equation is written in vertex form

y = a(x-h)^2 + k  where (h,k) is the vertex

y = 3(x- -19)^2 - 14

The vertex is (-19,-14)

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What value of x satisfies 3x + 2 = 14?
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Hey! 
Your answer would be 4! 
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4 years ago
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Find the value of x.<br> B<br> 10-X<br> 3<br> D<br> х<br> 2.<br> A<br> С<br> x = [?]
Nuetrik [128]

Answer:

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7 0
3 years ago
Hello! Verify the identity. Please show your work! Use trigonometric identities to verify each expression is equal.
RSB [31]

Answer:

See Below.

Step-by-step explanation:

We want to verify the identity:

\displaystyle \csc^2 x -2\csc x \cot x +\cot ^2 x = \tan^2\left(\frac{x}{2}\right)

Note that the left-hand side is a perfect square trinomial pattern. Namely:

a^2-2ab+b^2=(a-b)^2

If we let <em>a</em> = csc(x) and <em>b</em> = cot(x), we can factor it as such:

\displaystyle (\csc x - \cot x)^2 = \tan^2\left(\frac{x}{2}\right)

Let csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x):

\displaystyle \left(\frac{1}{\sin x}-\frac{\cos x }{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)

Combine fractions:

\displaystyle \left(\frac{1-\cos x}{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)

Square (but do not simplify yet):

\displaystyle \frac{(1-\cos x)^2}{\sin ^2x}=\tan^2\left(\frac{x}{2}\right)

Now, we can make a substitution. Let <em>u</em> = <em>x</em> / 2. So, <em>x</em> = 2<em>u</em>. Substitute:

\displaystyle \frac{(1-\cos 2u)^2}{\sin ^22u}=\tan^2u

Recall that cos(2u) = 1 - sin²(u). Hence:

\displaystyle \frac{(1-(1-2\sin^2u))^2}{\sin ^2 2u}=\tan^2u

Simplify:

\displaystyle \frac{4\sin^4 u}{\sin ^2 2u}=\tan^2 u

Recall that sin(2u) = 2sin(u)cos(u). Hence:

\displaystyle \frac{4\sin^4 u}{(2\sin u\cos u)^2}=\tan^2 u

Square:

\displaystyle \frac{4\sin^4 u}{4\sin^2 u\cos ^2u}=\tan^2 u

Cancel:

\displaystyle \frac{\sin ^2 u}{\cos ^2 u}=\tan ^2 u

Since sin(u) / cos(u) = tan(u):

\displaystyle \left(\frac{\sin u}{\cos u}\right)^2=\tan^2u=\tan^2u

We can substitute <em>u</em> back for <em>x</em> / 2:

\displaystyle \tan^2\left(\frac{x}{2}\right)= \tan^2\left(\frac{x}{2}\right)

Hence proven.

3 0
3 years ago
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