To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)
Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)
FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>
Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em>
Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>
Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2
Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.
The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2
Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
Answer:
Step-by-step explanation:
let's start by separating the fraction into two new smaller fractions
.
First,<em> s(s^2+s+1)</em> must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.
Where <em>A</em>, <em>B</em> and <em>C</em> are unknown constants. The numerator of <em>s</em> is a constant <em>A</em>, because <em>s</em> is linear, the numerator of <em>s^2+s+1</em> is a linear expression <em>Bs+C</em> because <em>s^2+s+1</em> is a quadratic expression.
Multiply both sides by the complete denominator:
![[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]](https://tex.z-dn.net/?f=%5B%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%5D%5Cfrac%7Bs%2B1%7D%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%7D%3D%5B%5Cfrac%7BA%7D%7Bs%7D%2B%5Cfrac%7BBs%2BC%7D%7Bs%5E%7B2%7D%2Bs%2B1%7D%5D%5B%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%5D)
Simplify, reorganize and compare every coefficient both sides:
Solving the system, we find <em>A=1</em>, <em>B=-1</em>, <em>C=0</em>. Now:
Then, we can solve the inverse Laplace transform with simplified expressions:
The first inverse Laplace transform has the formula:
For:
We have the formulas:
We have to factorize the denominator:
It means that:
So <em>a=-1/2</em> and <em>b=(√3)/2</em>. Then:
![\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}=e^{-\frac{t}{2}}[cos\frac{\sqrt{3}t }{2}]\\\\\\\frac{1}{2}[\frac{2}{\sqrt{3} } ]\mathcal{L}^{-1}\{\frac{\sqrt{3}/2 }{(s+1/2)^{2}+3/4}\}=\frac{1}{\sqrt{3} } e^{-\frac{t}{2}}[sin\frac{\sqrt{3}t }{2}]](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5C%7B-%5Cfrac%7Bs%2B1%2F2%7D%7B%28s%2B1%2F2%29%5E%7B2%7D%2B3%2F4%7D%5C%7D%3De%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%5Bcos%5Cfrac%7B%5Csqrt%7B3%7Dt%20%7D%7B2%7D%5D%5C%5C%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%20%7D%20%5D%5Cmathcal%7BL%7D%5E%7B-1%7D%5C%7B%5Cfrac%7B%5Csqrt%7B3%7D%2F2%20%7D%7B%28s%2B1%2F2%29%5E%7B2%7D%2B3%2F4%7D%5C%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%20%7D%20e%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%5Bsin%5Cfrac%7B%5Csqrt%7B3%7Dt%20%7D%7B2%7D%5D)
Finally:
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.
Answer:
hahah 0x
Step-by-step explanation:
Given:
Nora has a day off every 4 day while Amirah has day off every 6 days .
Nora last day off was on 29 april while Amirah was on 1 may.
To find:
When they have same day off.
Solution:
Nora last day off was on 29 april and she has a day off every 4 day.
So, her off days are
29 april, 3 may, 7 may, 11 may, 14 may, ...
Amirah last day off was on 1 may and she has day off every 6 days .
So, her off days are
1 may, 7 may, 13 may, ...
From the above two lists of off days, it is clear that they have same day off on 7 may.
