Check the picture below.
so, that'd be 20.24845673131658693325, and to the nearest tenth it'd then be 20.2
so, that'd be 20.24845673131658693325, and to the nearest tenth it'd then be 20.2
How many area codes (ABC) would be possible if the first 2 digits can be any number 2-9 & last digit can be any number 0-9?
1 answer:
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There's no if about it,
has a zero
so 
is a factor. That's the special case of the Remainder Theorem; since we'll get a remainder of zero when we divide 
by 
At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover
too, so 
is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.
So far we have
where
is the zero we haven't guessed yet. Again we could divide by 
but just looking at the constant term we must have
so
We check
We usually talk about the zeros of a function and the roots of an equation; here we have a function whose zeros are
has a zero
At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover
So far we have
where
so
We check
We usually talk about the zeros of a function and the roots of an equation; here we have a function