 # Let a, b and c be a real numbers such that a>0, b<0 and c<0. Find the sign of each expression.a) -ab) bcc) a-bd) ab+ac sodni3 2021-03-09 Answered

Let a, b and c be a real numbers such that $a>0,b<0$ and $c<0$. Find the sign of each expression.
a) -a
b) bc
c) a-b
d) $ab+ac$

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(a) Here $a>0$ $⇒$ −a will be a negative number.
Thus, sign of -a is negative .
(b) It is known that product of two negative number will be positive.
As $b<0$ and $c<0$ $⇒$ $bc>0$.
Thus, sign of bc is positive.
(c) It is known that when a negative number is subtracted from a positive number, the resultant number will be positive.
Here $a>0,b<0$ $⇒$ $a-b>0.$
Thus, sign of a-b is positive.
(d) Here
$a>0,b<0$ $⇒$ $ab<0$
$a>0,c<0$ $⇒$ $ac<0$
So ac and bc negative.
It is known that sum of two negative number will be negative.
Thus, $ab+ac<0.$
Thus, sign of $ab+ac$ is .