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(B) Given,$|a-b|=2$,$|b-c|=3$,and $|c-d|=4$.We can write $a-d = (a-b) + (b-c) + (c-d)$.Since $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$,the possible values for

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$18$

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(B) Given,$|a-b|=2$,$|b-c|=3$,and $|c-d|=4$.We can write $a-d = (a-b) + (b-c) + (c-d)$.Since $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$,the possible values for $(a-b)$,$(b-c)$,and $(c-d)$ are $\pm 2$,$\pm 3$,and $\pm 4$ respectively.The possible values for $a-d$ are obtained by summing these combinations:$2+3+4 = 9$$2+3-4 = 1$$2-3+4 = 3$$2-3-4 = -5$$-2+3+4 = 5$$-2+3-4 = -3$$-2-3+4 = -1$$-2-3-4 = -9$Thus,the possible values for $|a-d|$ are $|9|, |1|, |3|, |-5|, |5|, |-3|, |-1|, |-9|$,which simplifies to the set $\{9, 5, 3, 1\}$.The sum of all possible values of $|a-d|$ is $9 + 5 + 3 + 1 = 18$.

Let $a, b, c, d$ be real numbers such that $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$. Then,the sum of all possible values of $|a-d|$ is

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