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Aleks04 [339]
2 years ago
14

Find the quotient of 21x^2y^6+6xy^3-30xy/3xy.

Mathematics
2 answers:
maks197457 [2]2 years ago
7 0

Answer:

Step-by-step explanation:

It’s c

kaheart [24]2 years ago
3 0
\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------

\bf \cfrac{21x^2y^6+6xy^3-30xy}{3xy}\implies \stackrel{\textit{distributing the denominator}}{\cfrac{21x^2y^6}{3xy}+\cfrac{6xy^3}{3xy}-\cfrac{30xy}{3xy}}
\\\\\\
\cfrac{21}{3}\cdot x^2x^{-1}y^6y^{-1}+\cfrac{6}{3}\cdot x^1x^{-1}y^3y^{-1}-\cfrac{30}{3}\cdot x^1x^{-1}y^1y^{-1}
\\\\\\
7x^{2-1}y^{6-1}+2x^{1-1}y^{3-1}-10x^{1-1}y^{1-1}\implies 7xy^5+2y^2-10
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Identify the monomial function described as odd or even, and indicate whether a is positive or negative.
Igoryamba

Answer:

Case 1: As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) decreases.

It is an 'odd' function with 'positive' a.

Case 2: As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) decreases.

It is an 'even' function with 'negative' a.

Case 3: As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) increases.

It is an 'even' function with 'positive' a.

Case 4: As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) increases.

It is an 'odd' function with 'negative' a.

Step-by-step explanation:

Let us consider a monomial function:

             f(x) = axⁿ

Case 1:

As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) decreases.

This happens only if a is 'positive' and n is 'odd'.  So, it is an 'odd' function with 'positive' a.

Case 2:

As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) decreases.

This happens only if a is 'negative' and n is 'even'. So, it is an 'even' function with 'negative' a.

Case 3:

As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) increases.

This happens only if a is 'positive' and n is 'even'. So, it is an 'even' function with 'positive' a.

Case 4:

As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) increases.

This happens only if a is 'negative' and n is 'odd'. So, it is an 'odd' function with 'negative' a.

Keywords: monomial function, odd function, even function

Learn more about monomial function from brainly.com/question/8973176

#learnwithBrainly

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Step-by-step explanation:

During the problem, secx = -5/2, we can assume that as cos = -2/5. -2 = x. 5 = r. find for Y with: x^2+y^2=r^2. After that, plug in for the variables and you get all the answers. Rationalize the square roots, don't forget.

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