Answer:
x= -2y/3
-4y+4y=14
0y= 14
This has no solution.
Step-by-step explanation:
Answer In cases of two equations having two unknown variables the simultaneous method of solving the equations is adopted.
Supposing “0” as the value of “k” we will have the equation as 3x+2y=0
Then separating variable x will give, x= -2y/3
Substituting the value of x into the first equation that is: 6x + 4y = 14
6(-2y/3)+4y=14
Expanding this we get: 6(-2y/3)+4y=14
-4y+4y=14
0y= 14
Thus, no solution.
Answer:
20 students
Step-by-step explanation:
If the class decreased by 15%, the students that she has now (17) represents a percentaje of:
100% - 15% = 85%
so<u> the 17 students are 85% of what she had</u>:
Students Percentage
17 ⇒ 85%
and we are looking for how many students she had 2 years ago, thus we are looking for the <u>100%</u> of students (the original number of studens). If we represent this number by x:
Students Percentage
17 ⇒ 85%
x ⇒ 100%
and we solve this problem using the <u>rule of three</u>: multiply the cross quantities on the table( 17 and 100) and then divide by the remaining amount (85):
x = 17*100/85
x = 1700/85
x=20
2 years ago she had 20 students
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).
