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Mathematics

2 answers:

faltersainse [42]3 years ago

3 0

An arithmetic sequence is a sequence where you could add or subtract as the patter stays the same. So, the answer would be (C.) or I and IV because if you take a look at them, for I you add 4 each time and for IV, you add 7 each time. FYI, II is NOT an arithmetic sequence and rather a geometric because every time, the constant is multiplied and not added or subtracted.

saul85 [17]3 years ago

3 0

Answer:

the answer would be (C.)

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

This is really confusing me:<br> Simplify so that your answers contain only positive exponents.

xxTIMURxx [149]

Answer:

see below

Step-by-step explanation:

For simplifying expressions of this sort, there are four rules of exponents that come into play;

$a^ba^c=a^{b+c}\\\\(ab)^c=a^cb^c\\\\(a^b)^c=a^{bc}\\\\\dfrac{1}{a^b}=a^{-b}\ \quad\text{which means}\quad a^b=\dfrac{1}{a^{-b}}$

I find it convenient to eliminate the fractions by adding the exponents, then rewrite any negative exponents as denominator factors.

$23. \quad\dfrac{1}{x^{-2}}=x^2\\\\24. \quad\dfrac{mn}{m^2n^3}=m^{1-2}n^{1-3}=m^{-1}n^{-2}=\dfrac{1}{mn^2}\\\\25. \quad\dfrac{k^{-2}}{k^{-3}}=k^{-2-(-3)}=k\\\\26. \quad\left(\dfrac{m^2}{n^3}\right)^3=\dfrac{m^{2\cdot 3}}{n^{3\cdot 3}}=\dfrac{m^6}{n^9}\\\\27. \quad\dfrac{x^2y^3z^4}{x^{-3}y^{-4}z^{-5}}=x^{2-(-3)}y^{3-(-4)}z^{4-(-5)}=x^5y^7z^9$