Prove that if a and b are positive integers, then there exists a unique integers q and r such that a=bq+r where 0≤r<b​

Step-by-step explanation:

Correct option is

C

0≤r<b

If r must satisfy0≤r<b

Proof,

..,a−3b,a−2b,a−b,a,a+b,a+2b,a+3b,..

clearly it is an arithmetic progression with common difference b and it extends infinitely in both directions.

Let r be the smallest non-negative term of this arithmetic progression.Then,there exists a non-negative integer q such that,

a−bq=r

=>a=bq+r

As,r is the smallest non-negative integer satisfying the result.Therefore, 0≤r≤b

Thus, we have

a=bq1+r1,   0≤r1≤b

y=mx+b

step-by-step explanation:

m slope

b y-intercept

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation: