A gardener wishes to design a rectangular rose garden with one side of the garden along a main road. The fencing for the 3 non-r
2 answers:
Answer:
There is no minimum cost for the fencing that can satisfy these conditions.
as the equation obtained for this problem is y = .
Step-by-step explanation:
let length of the rectangular rose garden be x
and breath of the rectangular rose garden be y
then the area of the garden = xy
cost of fencing a foot for the 3 non road sides = $6
and cost of fencing a foot for a road side = $8
total cost of fencing = 6x + 8x + 6y + 6y
so, total cost = 14x + 12y
cost of fencing for every square foot of the area =$2
So, 2xy = 14x + 12y
=> 2xy = 14x + 12y
Also, x and y has to satisfy x>0 and y>0
You can solve for y (or x if you prefer)
2xy-12y=14x
xy - 6y = 7x
y(x-6)=7x
y =
we can check that this function has no minimum value.
as we increase x (starting any x > 6 to make y positive) to make y minimal, y will decrease but the product xy will increase.
So, there is not a minimum cost for fencing that can satisfy these conditions.
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1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
The number 1052, written as a base 6 number is 4512
Given: 1052 toothpicks
To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number
First we note that,
This implies that exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.
Now, and
. This implies that
exceeds 1052 and thus there can be at most 4 groups of
.
Then,
So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.
Now, and
. This implies that
exceeds 188 and thus there can be at most 5 groups of
.
Then,
So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.
Now, and
. This implies that
exceeds 8 and thus there can be at most 1 group of
.
Then,
So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.
Now, and
. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.
This concludes the grouping.
Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
Then,
So, the number 1052, written as a base 6 number is 4512.
Learn more about change of base of numbers here:
Answer:
brownies per hour
Step-by-step explanation:
Given parameters:
Number of brownies = of a dozen
Time taken = 3hrs
Unknown:
Average baking per hour = ?
Solution:
The average baking per hour;
This is a rate problem;
Average baking per hour =
Number of brownies = x 12 =
Average baking per hour =
=
=
= 3brownies per hour