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3 years ago

Amelia needs 3 2/3 cups of sugar to make cupcakes. Which improper fraction names this amount?

Mathematics

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

3 2/3=

3*3+2=
9/3+2/3=
11/3

C. 11/3 cups

Good Studies!!

Dmitrij [34]3 years ago

3 0

C because you multiply the big number with the lower number then you add the number on top. 3 times 3 equals 9 plus 2 = 11. you keep the bottom number under your new number

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Use set notation to write the members of the following set, or state that the set has no members. Odd numbers between 2 and 82 t

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Let the set of all Odd multiples of 9 between 2 and 82 be denoted by D, then, using set-builder notation,

$D=\{ 18n+9 \mid n \in \mathbb{N}, 0\le n\le 4 \}$

The odd multiples of 9, $m$ , in the range $2\le m \le 82$ form the set

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Each member of the set is a term of the arithmetic progression

$U_n=18n+9$

where the values of $n$ range from 0 to 4, or $0\le n\le 4$

Putting these facts together, we get the result

$D=\{ 18n+9 \mid n \in \mathbb{N}, 0\le n\le 4 \}$

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Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer