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2 years ago

I saved the best for last :]

Mathematics

1 answer:

harkovskaia [24]2 years ago

4 0

Answer:

a¹⁸b²⁴c⁵⁴

Step-by-step explanation:

(aᵇ)ᶜ = aᵇᶜ = aᵇ×ᶜ

(-aⁿ)ᵒᵈᵈ = -aⁿ

(-aⁿ)ᵉᵛᵉⁿ = aⁿ

aⁿ × aⁿ = aⁿ⁺ⁿ

_________

In this situation, 6 is an even exponent which means that the negative coefficient of this 9th degree monomial will be positive. Now the only step is to multiply the exponents by the exponent outside of the perenthesis which is 6.

so (-a³b⁴c⁹)⁶ = ((-a³)⁶(b⁴)⁶(c⁹)⁶) = ((a³ˣ⁶)(b⁴ˣ⁶)(c⁹ˣ⁶)) = ((a¹⁸)(b²⁴)(c⁵⁴)) = a¹⁸b²⁴c⁵⁴

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Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

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Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .