Wednesday, April 15, 2020 (in North America and South America)
Thursday, April 16, 2020 (outside of North American and South America)
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The cost of 12 bags of avocados is
Thus, the chef spent
Solution 1
A bag of avocados sells for $5.00, and so a 10% discount represents a savings of
Thus, the discounted price for a bag of avocados is
A box of mangoes sells for $12.50, and so a 20% discount represents a savings of
Thus, the discounted price for a box of mangoes is
Solution 2
A bag of avocados sells for $5.00, and so a 10% discount is equivalent to paying
Thus, the discounted price for a bag of avocados is
A box of mangoes sells for $12.50, and so a 20% discount is equivalent to paying
Thus, the discounted price for a box mangoes is
Avocados are sold in bags of 6 and the chef needs 100 avocados.
Since
Mangoes are sold in boxes of 15 and the chef needs 70 mangoes.
Since
Avocados are sold for $5.00 per bag, and so the cost to purchase any number of bags is a whole number.
Mangoes are sold for $12.50 per box, and so the cost to purchase mangoes is a whole number only when an even number of boxes are bought (1 box costs $12.50, 2 boxes cost $25.00, 3 boxes cost $37.50, and so on).
Since the chef spends exactly $75.00 (a whole number), then the chef must purchase an even number of boxes of mangoes.
If the chef purchases 2 boxes of mangoes, the cost is $25.00, which leaves
In this case, the chef has
Each tart requires 1 avocado and 2 mangoes and so the chef can make
If the chef purchases 4 boxes of mangoes, the cost is
In this case, the chef has
Each tart requires 1 avocado and 2 mangoes and so the chef can make 30 tarts.
If the chef purchases 6 boxes of mangoes, the cost is
Purchasing more than 6 boxes of mangoes will cost the chef more than $75.00. Therefore, if the chef purchases 30 avocados (5 bags) and 60 mangoes (4 boxes), she will have spent exactly $75.00, have twice as many mangoes as avocados, and be able to make the greatest number of tarts, 30.
The parabola
A third vertex of the parabolic rectangle lies on the
If one vertex of a parabolic rectangle is
The vertex that lies vertically above
This vertex lies on the parabola
The width of the rectangle is equal to this
Let a vertex of the parabolic rectangle be the point
A second vertex (also on the
The width of this rectangle is given by the
The area of a parabolic rectangle having length
If such a parabolic rectangle has length 36, then
The area of this rectangle is
If such a parabolic rectangle has width 36, then
The area of this rectangle is
Let a vertex of the parabolic rectangle be the point
A second vertex (also on the
The width of this rectangle is given by the
The area of a parabolic rectangle having length
If the length and width of such a parabolic rectangle are equal, then
For
We begin by drawing triangulations to determine the values of
Although we would obtain these same four answers by positioning the interior points in different locations, or by completing the triangulations in different ways, the diagrams above were created to help visualize a pattern.
From the answers shown, we see that
We must justify why this observation is true for all
Notice that each triangulation (after the first) was created by placing a new interior point inside the previous triangulation.
Further, each square is divided into triangles, and so each new interior point is placed inside a triangle of the previous triangulation (since no 3 points may lie on the same line).
For example, in the following diagrams, we observe that
Also, each of the triangles outside of
Triangle
To triangulate the region defined by triangle
Thus, the placement of
That is,
the
specifically, the
this triangle contributed 1 to the value of
after the
this is a net increase of 2 triangles, and thus
Thus,
In the triangulation of a regular
All such triangulations of a regular
The reasoning used in part (b) extends to any regular polygon having
That is, each additional interior point that is added to the triangulation for
Thus,
So then
Solution 1
If
If
Thus if
If
If
Thus if
Therefore, for all possible values of
Solution 2
Using the definition twice and simplifying, we get
Since
If
If
Thus,
Solution 3
Using the definition twice and simplifying, we getSince
Thus, the product
Similarly,
Therefore, the product
Since
An integer is divisible by 10 exactly when its units (ones) digit is 0.
The difference
Thus, we will show that for all possible values of
When a non-negative integer is divided by 10, the remainder is one of the integers from 0 through 9, inclusive.
Thus for every possible choice for
If for example
Since
For example, if the units digit of
More generally, if the units digit of
(Can you explain why this is true?)
For example, if we know that
Given that we know all possible units digits of
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |
1 | 2 | 5 | 0 | 7 | 6 | 7 | 0 | 5 | 2 | |
2 | 5 | 6 | 1 | 0 | 7 | 0 | 1 | 6 | 5 | |
5 | 6 | 7 | 2 | 1 | 0 | 1 | 2 | 7 | 6 | |
6 | 7 | 0 | 5 | 2 | 1 | 2 | 5 | 0 | 7 | |
7 | 0 | 1 | 6 | 5 | 2 | 5 | 6 | 1 | 0 | |
0 | 1 | 2 | 7 | 6 | 5 | 6 | 7 | 2 | 1 | |
1 | 2 | 5 | 0 | 7 | 6 | 7 | 0 | 5 | 2 |
Looking at the table, we see that for each of the possible units digits for
Thus beginning at
Since
Further,
Every
Suppose that
If
If
Each possible choice of
If
If
Therefore,
Therefore,
Thus, we need to only determine when
For which of the possible values of
Every
From this table, we make the following observations:
if
if
if
if
if
Using these observations, we summarize the remainders of
0 | 1 | 2 | 3 | 4 | |
1 | 2 | 0 | 0 | 2 | |
2 | 0 | 1 | 1 | 0 | |
0 | 1 | 2 | 2 | 1 | |
1 | 2 | 0 | 0 | 2 |
With respect to division by 5, we see in the table above that for each of the possible remainders for
Thus beginning at
Since
Finally, we want to determine for which of the possible values of
Every
From this table, we make the following observations:
if
if
if
if
if
if
if
Using these observations, we summarize the remainders of
0 | 1 | 2 | 3 | 4 | 5 | 6 | |
1 | 2 | 5 | 3 | 3 | 5 | 2 | |
2 | 5 | 5 | 3 | 3 | 5 | 5 | |
5 | 5 | 5 | 3 | 3 | 5 | 5 | |
5 | 5 | 5 | 3 | 3 | 5 | 5 |
Looking at the table, we see that if
Also, if
Therefore,
Summary:
The values of
Thus, there are 10 possible values for