Question

a flower shop just received a shipment of flowers. the table shows the number of each type.
the florist selects from these flowers to make arrangements for the upcoming flower show.
use the drop - down menus to complete the statements.
the greatest number of identical arrangements that can be made using only the carnations and asters with no flowers left over is ▲
the 9 ▲ arrangements that can be made with no flower

Explanation:

Step1: Find the GCF of 42 and 72

• Prime factorize 42: \(42 = 2\times3\times7\)
• Prime factorize 72: \(72=2^{3}\times3^{2}\)
• The common factors are \(2\) and \(3\).
• GCF\((42,72)=2\times3 = 6\)

Step2: Calculate the number of arrangements and remaining flowers

• Number of arrangements using asters and carnations:
• For asters (\(n_1 = 42\)): \(\frac{42}{6}=7\)
• For carnations (\(n_2=72\)): \(\frac{72}{6} = 12\)
• Remaining flowers:
• Let's assume we are making arrangements with only asters and carnations.
• We are not using all the flower types in this part (the problem says "using only the carnations and asters").
• The total number of flowers used in arrangements: \(6\times(7 + 12)=6\times19=114\)
• But if we consider the formula for the number of remaining flowers when making arrangements with GCF \(d\) for two - number \(a\) and \(b\) (here \(a = 42\), \(b = 72\)), we can also use the property that if \(a=md\) and \(b=nd\) (\(m = 7\), \(n = 12\), \(d = 6\)), there is no remainder when dividing \(a\) and \(b\) by \(d\)

Answer:

The greatest number of identical arrangements that can be made using only the carnations and asters with no flowers left over is \(6\)