Answer:
Step-by-step explanation:
3(2x + 7) - 2 (x + 5)
need to distibute
5x+10-2x+3
combine like terms
3x+13
divide
x=4.3
For this case what you should know is that both functions are of the potential type.
We have then that
y = 2 * 2 ^ x This function grows exponentially upwards.
y = -2 * 5 ^ x This function grows exponentially downwards.
Answer See attached graphics.
The complete pattern is: 512, 256, 128, 64, 32, 16, 8, ( and extended ) 4, 2, 1!
I do this pattern in my head literally all the time!
we are given two points as
(3, 16) and (5, 10)
firstly , we will find slope
slope is m
so, we can use formula
now, we can plug points
we can plug it in y=mx+b
we get
now, we can select any one points
and find b
(3,16)
x=3 and y=16
we can find b
now, we can plug back b
and we get
so, equation is
and
...............Answer
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).
