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Please help me with parts b and c.

9514 1404 393

  b.  A × B = GCF(A, B) × LCM(A, B)

  c. 15 = GCF × LCM / 20

Step-by-step explanation:

b. The products of the numbers in the table are 80, 108, 216, 45.

The products of the corresponding GCF and LCM are 80, 108, 216, 45.

Conjecture: the product of the numbers is equal to the product of the GCF and LCM.

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c. GCF×LCM = 5×60 = 300 = A×B = 20×B

  B = 300/20 = 15

The other number is 15, found by using the conjecture of part b.

If you think for a minute about what the GCF and LCM are, you realize the relationship discussed in this problem must be the case. Consider the factors of two numbers A and B:

  A = (factors in common with B) × (factors unique to A)

  B = (factors in common with A) × (factors unique to B)

The GCF, by definition, are (factors in common between A and B).

The LCM, by definition, is the product of unique factors:

  (common factors) × (factors unique to A) × (factors unique to B)

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Then the product of the LCM and the GCF is ...

  GCF × LCM =

  (factors in common) × ((factors in common) × (unique to A) × (unique to B))

Using the associative and commutative properties of multiplication, we can rearrange this product to be ...

  ((factors in common)×(unique to A)) × ((factors in common)×(unique to B))

  = A × B

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Personally, I use a kind of diagram to represent the factorization of A and B and their LCM and GCF:

  [unique to A (common] unique to B)

Then A = factors in [ ], and B = factors in ( ). The stuff in (common] is the GCF, and the overall product is the LCM.

Using the example of part c, this would look like [4 (5] 3), so A = 4·5 = 20 and B = 5·3 = 15. The GCF is 5, and the LCM is 4·5·3 = 60.

The end parts [unique to A( and ]unique to B) can have no common factors. Any common factors must reside in the (common] part.

The answer is:

 b.  A × B = GCF(A, B) × LCM(A, B)

 c. 15 = GCF × LCM / 20

i believe the answer is d.

hope it you! : d

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: