What is the largest number in the form x987yz that is divisible by 12?

998796

Step-by-step explanation:

For a number to be divisible by 12. it must be divisible by 3 and  4

Since we need the largest 6 digit number, we will take x = 9

So the number is 9987yz

Again since we need the largest number we will consider the largest 2 digit number multiple of 4 and 3.

Largest 2 digit number divisible by 4 is 96, which is also divisible by 3.

So, yz  = 96

Therefore , the largest number in the form will be : 998796

The largest number is 998796

The largest digit you could use is 9 so, our number would be, 998799,

but the thing is that it's not divisible by 12.

The next largest would be 8. (998798) Not divisible.

7, so our number would be, 998797. Not divisible.

6, 998796. 998796/12= 83233, which means that it is divisble, so, the largest number in the form x987yz that is divisble by 12 is 998796

Please correct me if I'm wrong.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.