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faust18 [17]
3 years ago
15

Find three consecutive integers with a sum of 267

Mathematics
2 answers:
Tanya [424]3 years ago
6 0
88 + 89 + 90 = 267 and they are consecutive integers as they are right next to each other.

the easiest way to solve this problem is to divide 267/3 to get 89, and then use this information to gather your three integers that should be of about the same value.
Bad White [126]3 years ago
6 0
267 divided 3 equals 89 so the numbers average to 89.

So 88,89,90 are the only ones that average to 89 and add up to 267

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Let z1 = a1 + b1i, z2 = a2 + b2i, and z3 = a3 + b3i. Prove the folowing using algebra or by showing with vectors.
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a)z1 +z2 =z2 + z1 ...proved.

b) z1 + ( z2+ z3 )=(z1+z2)+z3 ... proved.

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}

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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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3 years ago
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