Tarush is a landscape architect. For his first public project he is asked to create a small scale drawing of a garden to be plac
ed in the corner of a city park. The garden is a right triangle with base 10\text{ m}10 m10, start text, space, m, end text and height 15\text{ m}15 m15, start text, space, m, end text. Draw the garden such that 111 unit on the grid below represents 2\text{ m}2 m2, start text, space, m, end text.
2 answers:
Answer:
See the attached figure 2.
Step-by-step explanation:
The question is as shown at the attached figure 1.
given: The garden is a right triangle with base 10 m and height 15 m.
We need to draw the garden such that 1 unit on the grid represents 2 m.
So the scale factor is 1 unit = 2 m
base = 10 m = 10/2 = 5 units
height = 15 m = 15/2 = 7.5 units
So, the garden on the grid will be a right triangle with base 5 units and height 7.5 units.
See the attached figure.
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Answer:
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the length side KJ
In the right triangle JKM
Applying the Pythagoras Theorem
we have
substitute
simplify
step 2
Find the value of cosine of angle MJK in the right triangle JKM
substitute the values
simplify
-----> equation A
step 3
Find the value of cosine of angle MJK in the right triangle JKL
we have
----> remember equation A
substitute the values
Simplify
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:
. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be
. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.