The florist can order roses bunches of one dozen and lilies in bunches of 8. Last month she ordered the same number of roses as lilies. If she ordered no
more than 100 roses, how many bunches of each could she have ordered?
What is the smallest number of bunches of each that she could have
ordered? Explain your reasoning.
2. Find the GCF from the two numbers, and rewrite the sum using the
distributive property.
A. 12.+18
B.42+14
C.36+27

9514 1404 393

  1. (rose bunches, lily bunches) ∈ {(2, 3), (4, 6), (6, 9), (8, 12)}. The least is (2, 3).

  2. 6(2+3); 14(3+1); 9(4+3)

Step-by-step explanation:

I like Euclid's algorithm for finding the greatest common factor (GCF): divide the larger number by the smaller, and note the remainder. If the remainder is zero, the smaller is the GCF; if not, use the remainder in place of the larger number and repeat.

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1. The GCF of 8 and 12 is 4, so the least common multiple is ...

  LCM(a, b) = ab/GCF(a, b)

  LCM(8, 12) = 8·12/4 = 24

24 roses is 2 bunches; 24 lilies is 3 bunches. Up to 8 bunches of roses can be ordered to keep the number less than 100.

The florist could have ordered these numbers of bunches (roses, lilies):

  (2, 3), (4, 6), (6, 9), (8, 12)

The least order is 2 bunches of roses and 3 bunches of lilies.

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2.

A. GCF = 6, so the factored form is 6(2+3)

B. GCF = 14, so the factored form is 14(3+1)

C. GCF = 9, so the factored form is 9(4+3)

  q

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )