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:
c!!
Step-by-step explanation:
last saturday i did a worksheet that had the sae question
<em><u>4 pints of strawberries and 2 pints of blueberries are bought</u></em>
<em><u>Solution:</u></em>
Let "a" be the pints of strawberries bought
Let "b" be the pints of blueberries cost
Cost per pint of strawberry = $ 1.60
Cost per pint of blueberry = $ 2.30
<em><u>A shopper bought twice as many pints of strawberries as pints of blueberries</u></em>
Therefore,
a = 2b --------- eqn 1
<em><u>They spent a total of $11.00. Therefore we frame a equation as:</u></em>
pints of strawberries bought x Cost per pint of strawberry + pints of blueberries cost x Cost per pint of blueberry = 11

1.6a + 2.3b = 11 --------- eqn 2
<em><u>Substitute eqn 1 in eqn 2</u></em>
1.6(2b) + 2.3b = 11
3.2b + 2.3b = 11
5.5b = 11
Divide both sides by 11
b = 2
<em><u>Substitute b = 2 in eqn 1</u></em>
a = 2(2)
a = 4
Thus 4 pints of strawberries and 2 pints of blueberries are bought
Answer: y = 4 - 3x/8
Step-by-step explanation: Subtract 3x from both sides of the equation and then divide by 8.
120 is good engough if you dont kn o