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IgorLugansk [536]
3 years ago
13

Prove: The sum of two multiples of 3 is a multiple of 3

Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

  3 . . . goes in both boxes

Step-by-step explanation:

Use the distributive property. Remove the common factor to outside parentheses:

  3(m +n) = a multiple of 3

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Insert the parenthesis to make the equality right:<br><br><br> 180÷300–30x9+199=205
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180÷(300-30)×9+199
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Given that p²,q² are two roots of x²-x+16=0. <br>Form another equation with roots 1/p,1/q.
statuscvo [17]

Answer:

  x² -3/4x +1/4 = 0

Step-by-step explanation:

Consider the two equations in factored and expanded forms:

  (x -p²)(x -q²) = x² -(p²+q²)x +p²q² = 0   ⇒   p²+q² = 1, p²q² = 16

and

  (x -1/p)(x -1/q) = x² -(1/p+1/q)x +1/(pq) = 0

Consider the squares of the sum and product of roots:

  constant term: (1/(pq))² = 1/(p²q²) = 1/16   ⇒   1/(pq) = √(1/16) = 1/4

  x-term: (1/p +1/q)² = (p +q)²/(pq)² = (p² +q² +2pq)/(p²q²)

  = (p² +q²)/(p²q²) +2/(pq)

  = 1/16 +2/√16 = 9/16   ⇒   (1/p +1/q) = √(9/16) = 3/4

Then the equation with roots 1/p and 1/q is ...

  x² -3/4x +1/4 = 0

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3 years ago
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Answer:

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Step-by-step explanation:

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