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bixtya [17]
2 years ago
12

Cindi is planting a rectangular flower bed with 40 orange flower and 28 yellow flowers. She wants to plant them so that each row

will have the same number of plants but of only one color. How many rows will Cindi need if she puts the greatest possible number of plants in each row?(show all work please)
Mathematics
1 answer:
Vilka [71]2 years ago
3 0

Answer: 17 rows, 7 of yellow flowers and 10 of orange flowers.

Step-by-step explanation:

We have 40 orange flowers

we have 28 yellow flowers.

The total number of flowers is 40 + 28= 68

Now, we want to find the maximal comon factor between 28 and 40 (that will be the number of flowers in each row)

for this, we took the smallest number and find its factors, and then we see if those factors are also of the greater number.

28/1 = 28 ----> 40/28 is not integer, so 28 is not a factor of 40.

28/2 = 14 ----> 40/14 is not integer, so 14 is not a factor of 40.

28/3 is not integer.

28/4 = 7 ----> 40/7 is not integer, so 7 is not a factor of 40.

28/5 is not integer.

28/6 is not integer.

28/7 = 4, 40/4 = 10, so 4 is a factor of both 28 and 40.

Then the maximum number of flowers that we can put in each row is 4.

this means that we have:

28/4 = 7 rows of yellow flowers (with 4 flowers each)

40/4 = 10 rows of orange flowers (with 4 flowers each)

a total of 17 rows.

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Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
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Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

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