Answer:
Area = 
x²
Step-by-step explanation:
The area (A) of a triangle is calculated as
A = 
bh ( b is the base and h the perpendicular height )
Here b = 
x and h = x, thus
A = × x × x = x²
You would need the distance formula
Insert the numbers
Solve inside
Square and add
Square root
4 units is the distance
To prove two sets are equal, you have to show they are both subsets of one another.
• <em>X</em> ∩ (⋃ ) = ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }
Let <em>x</em> ∈ <em>X</em> ∩ (⋃ ). Then <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ ⋃ . The latter means that <em>x</em> ∈ <em>S</em> for an arbitrary set <em>S</em> ∈ . So <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ <em>S</em>, meaning <em>x</em> ∈ <em>X</em> ∩ <em>S</em>. That is enough to say that <em>x</em> ∈ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }. So <em>X</em> ∩ (⋃ ) ⊆ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }.
For the other direction, the proof is essentially the reverse. Let <em>x</em> ∈ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }. Then <em>x</em> ∈ <em>X</em> ∩ <em>S</em> for some <em>S</em> ∈ , so that <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ <em>S</em>. Because <em>x</em> ∈ <em>S</em> and <em>S</em> ∈ , we have that <em>x</em> ∈ ⋃ , and so <em>x</em> ∈ <em>X</em> ∩ (⋃ ). So ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ } ⊆ <em>X</em> ∩ (⋃ ).
QED
• <em>X</em> ∪ (⋂ ) = ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }
Let <em>x</em> ∈ <em>X</em> ∪ (⋂ ). Then <em>x</em> ∈ <em>X</em> or <em>x</em> ∈ ⋂ . If <em>x</em> ∈ <em>X</em>, we're done because that would guarantee <em>x</em> ∈ <em>X</em> ∪ <em>S</em> for any set <em>S</em>, and hence <em>x</em> would belong to the intersection. If <em>x</em> ∈ ⋂ , then <em>x</em> ∈ <em>S</em> for all <em>S</em> ∈ , so that <em>x</em> ∈ <em>X</em> ∪ <em>S</em> for all <em>S</em>, and hence <em>x</em> is in the intersection. Therefore <em>X</em> ∪ (⋂ ) ⊆ ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }.
For the opposite direction, let <em>x</em> ∈ ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }. Then <em>x </em>∈ <em>X</em> ∪ <em>S</em> for all <em>S</em> ∈ . So <em>x</em> ∈ <em>X</em> or <em>x</em> ∈ <em>S</em> for all <em>S</em>. If <em>x</em> ∈ <em>X</em>, we're done. If <em>x</em> ∈ <em>S</em> for all <em>S</em> ∈ , then <em>x</em> ∈ ⋂ , and we're done. So ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ } ⊆ <em>X</em> ∪ (⋂ ).
QED
Answer:1st one 2nd one and 8th
Step-by-step explanation:
