Answer:
she has $78.5
Explanation:
1- She has 5 rolls of pennies containing 50 coins each.
Total pennies she has = 5*50 = 250 pennies
Now, we will get the amount of pennies she has in dollars as follows:
1 penny ............> 0.01 dollar
250 pennies .....> ??
Amount is dollars = 250*0.01 = $2.5 .............> I
2- She has 3 rolls of nickels containing 40 coins each
Total nickels = 3*40 = 120 nickels
Now, we will get the amount of nickels she has in dollars as follows:
1 nickel ..............> 0.05 dollar
120 nickels .........> ??
Amount in dollars = 120*0.05 = $6 ............> II
3- She has 6 rolls of dimes containing 50 coins each
Total dimes = 6*50 = 300 dimes
Now, we will get the amount of dimes she has in dollars as follows:
1 dime ...........> 0.1 dollar
300 dimes .....> ??
Amount in dollars = $30 ...............> III
4- She has 4 rolls of quarters containing 40 coins each.
Total quarters = 4*40 = 160 quarters
Now, we will get the amount of quarters she has in dollars as follows:
1 quarter ...........> 0.25 dollar
160 quarters ........> ??
Amount in dollars = $40 ..........> IV
5- Final step:
we will add I,II,III and IV to get the total amount of money she has as follows:
Total amount of money = 2.5 + 6 + 30 + 40
Total amount of money = $78.5
Hope this helps :)
Answer:
stream walls by louis tomlinson on itunes
Step-by-step explanation:
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:
He drank 2¹/₄qt of 6 cans
Step-by-step explanation:
A can of killa kola contains 3/8 qt.
Kyle drinks 6 cans.
asking how much he drank is multipling the quantity in a can by how many cans he drank
Here it is 3/8 x 6 = 18/8 = 9/4 = 2¹/₄qt
1 - 51
2 - 129
3 - 51
5 - 90
