Answer:
Step-by-step explanation:
y=x^2-x+1
We want to solve for x.
I'm going to use completing the square.
Subtract 1 on both sides:
y-1=x^2-x
Add (-1/2)^2 on both sides:
y-1+(-1/2)^2=x^2-x+(-1/2)^2
This allows me to write the right hand side as a square.
y-1+1/4=(x-1/2)^2
y-3/4=(x-1/2)^2
Now remember we are solving for x so now we square root both sides:
The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.
This is only the right side of the parabola because of the domain restriction. We want x-1/2 to be positive.
That is we want:
Add 1/2 on both sides:
The last step is to switch x and y:
Answer:
1:13
Step-by-step explanation:
A standard deck has 52 cards not including the joker. each suit has one 10. there are total four "10" in a deck of playing cards. 4/52 = 1/13 or 1:13
hopefully this will help :)
Answer:
Step-by-step explanation:
Given
Square
Triangle
Required
Find h, if both shapes have the same area
The area of the square is:
The area of the triangle is:
Equate both areas
Divide both sides by x
Multiply both sides by 2
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).
