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Morgarella [4.7K]
4 years ago
14

What is the solution set for the inequality x < -2 or x > 6

Mathematics
1 answer:
bija089 [108]4 years ago
8 0

he solution set is

{ x ∣ x > 1 } .


Explanation

For each of these inequalities, there will be a set of

x

-values that make them true. For example, it's pretty clear that large values of

x

(like 1,000) work for both, and negative values (like -1,000) will not work for either.

Since we're asked to solve a "this OR that" pair of inequalities, what we'd like to know are all the

x

-values that will work for at least one of them. To do this, we solve both inequalities for

x

, and then overlap the two solution set

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Answer:

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Step-by-step explanation:

It is given that there are three vectors z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3. Now, we have to prove (a) z1 + z2 = z2 + z1 and (b) z1 + (z2 +z3) = (z1 + z2) + z3.

(a) z1 + z2 = (a1 +ib1) + (a2+ ib2) = (a1 +a2) + i(b1 +b2) {Adding the real and imaginary parts separately}

Again, z2 + z1 =(a2 +ib2) + (a1 +ib1) = (a2 +a1) + i(b2 +b1) {Adding the real and imaginary parts separately}

Hence, z1 +z2 =z2 + z1 {Since, (a1 +a2) = (a2 +a1) and (b1 +b2) = (b2 +b1)}

(b) z1 + ( z2+ z3 ) = [a1 + ib1] + [(a2 + a3 ) + i(b2 + b3 )] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Again, (z1+z2)+z3 = [(a1+a2) +i(b1+b2)]+[a3+ib3] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Hence, z1 + ( z2+ z3 )=(z1+z2)+z3 proved.

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