Add the bases together and divide by 2 to get 14.
Notes:
The notation ">=" without quotes means "greater than or equal to"
The upper case "U" means "set union"
Instead of using the intersection symbol, I will use a lower case 'n'
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Problem 1
A = {x | x < 1} which is the set of x values smaller than 1
B = {x | x >= 5} is the set of x values that are equal to 5 or larger
A U B = set of values that are from set A OR they are from set B (or both)
A U B = {x | x < 1 or x >= 5}
we simply connect the two inequalities mentioned with an "or"
note: how there is no overlap between the two regions. The "U" means "set union" which is like a sort of glue to tie the two sets together with an "or".
Answer: {x| x < 1 or x >= 5}
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Problem 2
A = {x | x < 1}
C = {x | x = 5} which is the set of one value only: 5 (x cannot equal any other value)
A U C = {x| x < 1 or x = 5}
So if a number is in set A U C, then this number is either less than 1, OR it is equal to 5
Answer: {x|x < 1 or x = 5}
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Problem 3
B U C = {x | x >= 5} because set C already has the "or equal to" part in there.
Set C is a subset of set B. If an item is in set C, then it is also in set B.
Answer: {x| x >= 5}
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Problem 4
Again recall that I'm using an 'n' to indicate "set intersection" instead of the upside down "U" symbol
A n B is the set of items that are in BOTH sets A and B at the same time. From problem 1, I mentioned there's a gap. There is no x value that is both less than 1 AND greater than or equal to 5. So this means that
A n B = empty set
which we use the "O" with a slash through it. This is a special symbol to indicate "empty set"
Another way to write "empty set" is to use curly braces with nothing inside like so { }
Answer: The "O" with a slash through it
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Problem 5
B n C is the same as {x| x = 5}
Why? Because if an item is in B n C, then it has to be in BOTH set B and set C at the same time. The only way this happens is if x = 5. If x is any other value, then it won't be in set C
Answer: {x| x = 5}
V=
note, r and R are differnt
so multipl both sides by (r+R)
V(r+R)=12R
distribute
Vr+VR=12R
minus VR from both sides
Vr=12R-VR
undistribute R
Vr=R(12-V)
divide both sides by (12-V)
or
Answer:
7
Step-by-step explanation:
