3 years ago

Segment addition and midpoints

Mathematics

1 answer:

aleksley [76]3 years ago

3 0

Answer:

$16$

Step-by-step explanation:

HOPE IT HELPS

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Step-by-step explanation:

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3 years ago

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4 years ago

Read 2 more answers

What condition renders this true, where S and T are sets?

daser333 [38]

Answer:

There are no sets S, T for which

$S \cup T^c=S^c\cap T$

holds

Step-by-step explanation:

Let the set A be

$A=S\cup T^c$

By de De Morgan's Law

$A^c=S^c\cap ((T)^c)^c$

But

$((T)^c)^c=T$

$A^c=S^c\cap T$

We conclude that

$S \cup T^c=S^c\cap T\Rightarrow A=A^c$

which is a contradiction because no set is equal to its complement.

3 0

3 years ago