If you can divide 81 by 4 and end up with a remainder of 1, then yes, 81 is a part of your sequence.
To justify that, it helps to think about ways we can represent numbers in terms of quotients and remainders. For example, if we take 10/3, we’ll obtain the quotient 3 with a remainder of 1, so another way of writing 10 would be 3(3)+1. If we replace that second three with any integer n, 3n+1 represents *every number* that leaves us with a remainder of 1 when we divide it by 3.
By the same logic, 4n+1 represents every number that leaves a remainder of 1 when we divide it by 4. 81 indeed meets this requirement, since can be written as 4(20)+1, so it would be a part of the sequence.
B even know like ur gragh isnt big enough be best fits it c is the next best thing
Answer:
(1,-1)
Step-by-step explanation:
y = -1
5x+7y = -2
Substitute the first equation into the second
5x + 7(-1) =-2
5x -7 = -2
Add 7 to each side
5x-7+7 = -2+7
5x= 5
Divide by 5
5x/5 = 5/5
x =1
(1,-1)
Its 6. 4 x 6 = 24 so multiply 1 by 6 and thats your answer
Step-by-step explanation:
We are given that x < 13 so: x equals every number <u>below</u> 13. For the table:
x
9 YES because 9 < 13
11 YES because 11 < 13
15 NO because 15 is NOT less than 13
19 NO because 19 is NOT less than 13
