Which undefined geometric term is described as a two-dimensional set of points that has no beginning or end?

Mathematics

2 answers:

lilavasa [31]3 years ago

6 0

A two dimensional set would be a xy-plane with no boundaries.

Alchen [17]3 years ago

5 0

The answer is plane because it is a two dimensional figure

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Joe is preparing 18 hot dogs for his party. However, Joe only has 12 hot dog buns. How many more hot dog buns does Joe need

fredd [130]

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7 0

3 years ago

Mathematics 13 don’t understand

UNO [17]

Answer:

1/6^3

Step-by-step explanation:

The applicable rules of exponents are ...

a^-b = 1/a^b

(a^b)(a^c) = a^(b+c)

Your expression can be simplified as follows:

$\dfrac{6^{-5}}{6^{-2}}=\dfrac{1}{(6^{-2})(6^5)}=\dfrac{1}{6^{-2+5}}=\boxed{\dfrac{1}{6^3}}$

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<em>Additional comment</em>

If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.

Consider multiplication:

$(x\cdot x\cdot x)\cdot(x\cdot x)=x^3\cdot x^2=x^{3+2}=x^5\\\\(x\cdot x\cdot x)\cdot(x\cdot x\cdot x)=(x^3)^2=x^{3\cdot2}=x^6$

Consider division:

$\dfrac{x\cdot x\cdot x}{x\cdot x}=x\quad\Longleftrightarrow\quad\dfrac{x^3}{x^2}=x^{3-2}=x^1\\\\\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{1}{x}\quad\Longleftrightarrow\quad\dfrac{x^2}{x^3}=x^{2-3}=x^{-1}$