Exercise 2.7.2: Proofs by cases - even/odd integers and divisibility. About Prove each statement. (a) If is an integer, then is odd. Solution (b) If integers and have the same parity, then is even. The parity of a number tells whether the number is odd or even. If and have the same parity, they are either both even or both odd. (c) If integers and where are consecutive, then they have opposite parity. (d) For integers and , if is odd, then is odd and is odd.

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation:

just took the test and my answers down below!

step-by-step explanation:

check the pic :

there is only one way to arrange all the eight relative to each other because they are not distinct. we have

now there are nine places to place an (and not more than one in one place).   because we have four symbols, we need to choose four of these nine places to insert an . order does not matter because the are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

thus there are 126 ways to form such "words."