Nobody will be able to give an accurate response to this question without seeing the tally chart sorry
25 kids in all
2xboys:3x girls
so the equation would be
2x+3x=25
solve,
5x=25
/5 /5
x=5
2x=10
3x=15
10 boys and 15 girls
The answer would be 0 solutions.
Here, we see <em>|</em><em />x+6<em>|</em><em /> = 2.
Oh wow! A foreign object!
|x+6|... two lines... what is that?
That is called absolute value. Whatever is inside the two lines, must have a positive answer!
Let's pretend we have a machine that has this absolute value function activated.
What we put in, we must get a positive answer out.
Let's put in -6.
-6 ==> BEEP BEEP ==> 6
Let's try 3.
3 ==> BEEP BEEP ==>3
Whatever we put in, if it is negative or positive, what comes out is always positive.
So, for how many values <em>x</em> is |x+6|=-2 true?
None, because the answer <em>must</em><em /> be positive!
-2 is not positive, <em>2</em><em /> is.
Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
