I3/I6(8O)= 65m of electrical cable left
Well, you could assign a letter to each piece of luggage like so...
A, B, C, D, E, F, G
What you could then do is set it against a table (a configuration table to be precise) with the same letters, and repeat the process again. If the order of these pieces of luggage also has to be taken into account, you'll end up with more configurations.
My answer and workings are below...
35 arrangements without order taken into consideration, because there are 35 ways in which to select 3 objects from the 7 objects.
210 arrangements (35 x 6) when order is taken into consideration.
*There are 6 ways to configure 3 letters.
Alternative way to solve the problem...
Produce Pascal's triangle. If you want to know how many ways in which you can choose 3 objects from 7, select (7 3) in Pascal's triangle which is equal to 35. Now, there are 6 ways in which to configure 3 objects if you are concerned about order.
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).
Answer:
Step-by-step explanation:


Answer:
Part A:
The graph passes through (0,2) (1,3) (2,4).
If the graph that passes through these points represents a linear function, then the slope must be the same for any two given points. Using (0,2) and (1,3). Write in slope-intercept form, y=mx+b. y=x+2
Using (0,2) and (2,4). Write in slope-intercept form, y=mx+b. y=x+2. They are the same and in graph form, it gives us a straight line.
Since the slope is constant (the same) everywhere, the function is linear.
Part B:
A linear function is of the form y=mx+b where m is the slope and b is the y-intercept.
An example is y=2x-3
A linear function can also be of the form ax+by=c where a, b and c are constants. An example is 2x + 4y= 3
A non-linear function contains at least one of the following,
*Product of x and y
*Trigonometric function
*Exponential functions
*Logarithmic functions
*A degree which is not equal to 1 or 0.
An example is...xy= 1 or y= sqrt. x
An example of a linear function is 1/3x = y - 3
An example of a non-linear function is y= 2/3x