what is the value when (n-2)^2+n-1when n=4?

Answer 1

To find the value of (n-2)^2 + n-1 when n=4, we simply substitute 4 for n in the expression:

(n-2)^2 + n-1

When we do this, the expression becomes:

(4-2)^2 + 4-1

Now we need to simplify the expression within the parentheses first. The expression within the parentheses is (4-2)^2, which means we need to subtract 2 from 4 and then square the result. So, we get:

2^2

Which equals 4. Now we can substitute this value back into the original expression to get:

4 + 4-1

Next, we simplify the expression 4-1, which equals 3. So, the expression becomes:

4 + 3

And we can simplify that further to get:

7

So, when n=4, the value of (n-2)^2 + n-1 is 7