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lesantik [10]
2 years ago
10

Xby the power of 5 times y to the power of 6 over 2 by the power of -2 times xby the power of 0times x by the power of 9

Mathematics
1 answer:
mars1129 [50]2 years ago
8 0

Answer:

We have the sentence:

"X by the power of 5 times y to the power of 6 over 2 by the power of -2 times x by the power of 0times x by the power of 9"

Let's break it into parts.

"X by the power of 5 times y to the power of 6..."

This can be written as:

x^5*y^6

"...  2 by the power of -2 times x by the power of 0times x by the power of 9"

This can be written as:

2^(-2)*x^(0)*x^(9)

And we have the quotient between the first thing and the second thing, then the equation is:

\frac{x^5*y^6}{2^{-2}*x^0*x^9}

And any number by the power of 0 is equal to 1, then:

x^0 = 1, then we can rewrite the equation as:

\frac{x^5*y^6}{2^{-2}*x^9}

We can keep simplifying this.

We know that:

a^(-n) = (1/a)^(n)

Then:

2^(-2) = (1/2)^2 = 1/4

Then we get:

\frac{x^5*y^6}{2^{-2}*x^9} = \frac{x^5*y^6}{x^9}*4

And we also know that:

a^n/a^m = a^(n - m)

Then:

\frac{x^5*y^6}{x^9}*4 = 4*y^6*\frac{x^5}{x^9} = 4*y^6*x^{5 - 9} = 4*y^6*x^{-4} = \frac{4*y^6}{x^4}

And we can't simplify this anymore.

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

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I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

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1. -35v + 70 - 7v

Negative multiplied by negative is a positive

2. -35v - 7v + 70

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3. -42v +70

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