<h3>
Answer: Choice C) 421.9</h3>
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Explanation:
You're on the right track. You wrote down the proper expression to get the final answer, assuming you meant to write 75/4 as the third term inside the parenthesis. This works because each time you cut the side length in half to get each smaller triangle's side. The 3 is because there are 3 sides for each of the triangles. Much of this I have a feeling you already know as you wrote down the expression on the page, though I'm not 100% sure of your mindset. Computing this expression leads to 421.875 which rounds to 421.9
note: an alternative is to write 3*75 + 3*75/2 + 3*75/4 + 3*75/8, though that is more work. It's better to have that 3 factored out.
The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.
3. We can see that
1+100=101
2+99=101
3+98=101
.................
and so on
From 1 to 100, there are 50 pairs that has the sum is 101
Z=101×50=5050
4. We can see that
1-2=-1
3-4=-1
5-6=-1
...........
and so on
There are 10 pairs that has sum is -1
Z= 10×(-1)=-10
Let's say "n" is a natural number. {1,2,3,4,..} To ensure we have an even number we will multiply "n" by 2. Two times any number will make an even number.
consecutive even numbers are like; 2, 4, 6, 8, 10 .. etc. Add +2 to the previous number to get the next consecutive.
1st even number = 2n
2nd even number = 2n + 2
3rd even number = 2n + 4
twice the first number (2n) is 20 more then the second (2n + 2).
2(2n) = 2n + 2 + 20
4n = 2n + 22
4n - 2n = 22
2n = 22
n = 11
Now use n = 11 to find the 3 consecutive even numbers.
1st even number = 2(11) = 22
2nd even number = 2(11) + 2 = 24
3rd even number = 2(11) + 4 = 26
22, 24, 26
