question number 1 is 0.6 and 1.4 number 2 is 0.2 and 2.5 number 3 is 20.0
and 26.1 number 4 is 6.54 and cannot see number 5 is 29.15 and cannot see number 6 is 32.10 and cannot see number 7 is 20.051 and cannot see number 8 is 36.582 and cannot see number 9 is 16.570 and cannot see thank you
Answer:
y(s) = ![\frac{5s-53}{s^{2} - 10s + 26}](https://tex.z-dn.net/?f=%5Cfrac%7B5s-53%7D%7Bs%5E%7B2%7D%20-%2010s%20%20%2B%2026%7D)
we will compare the denominator to the form ![(s-a)^{2} +\beta ^{2}](https://tex.z-dn.net/?f=%28s-a%29%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D)
![s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D%20-10s%2B26%20%3D%20%28s-a%29%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D%20%3D%20s%5E%7B2%7D%20-2as%20%2Ba%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D)
comparing coefficients of terms in s
1
s: -2a = -10
a = -2/-10
a = 1/5
constant: ![a^{2}+\beta ^{2} = 26](https://tex.z-dn.net/?f=a%5E%7B2%7D%2B%5Cbeta%20%5E%7B2%7D%20%3D%2026)
![(\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B5%7D%20%29%5E%7B2%7D%20%2B%20%5Cbeta%20%5E%7B2%7D%20%3D%2026%5C%5C%5C%5C%5Cbeta%5E%7B2%7D%20%3D%2026%20-%20%5Cfrac%7B1%7D%7B10%7D%20%5C%5C%5C%5C%5Cbeta%20%3D%5Csqrt%7B26%20-%20%5Cfrac%7B1%7D%7B10%7D%7D%20%3D5.09)
hence the first answers are:
a = 1/5 = 0.2
β = 5.09
Given that y(s) = ![A(s-a)+B((s-a)^{2} +\beta ^{2} )](https://tex.z-dn.net/?f=A%28s-a%29%2BB%28%28s-a%29%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D%20%29)
we insert the values of a and β
= ![A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )](https://tex.z-dn.net/?f=A%28s-0.2%29%2BB%28%28s-0.2%29%5E%7B2%7D%20%2B%205.09%5E%7B2%7D%20%29)
to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A
5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)
-52 = B(26)
B = -52/26 = -2
to get A lets substitute s=0.4
5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)
-51 = 0.2A - 52.08
0.2A = -51 + 52.08
A = -1.08/0.2 = 5.4
<em>the constants are</em>
<em>a = 0.2</em>
<em>β = 5.09</em>
<em>A = 5.4</em>
<em>B = -2</em>
<em></em>
Step-by-step explanation:
- since the denominator has a complex root we compare with the standard form
![s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D%20-10s%2B26%20%3D%20%28s-a%29%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D%20%3D%20s%5E%7B2%7D%20-2as%20%2Ba%5E%7B2%7D%20%2B%5Cbeta%20%5E%7B2%7D)
- Expand and compare coefficients to obtain the values of a and <em>β </em>as shown above
- substitute the values gotten into the function
- Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
- after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above
Thanks...
Answer:
Dr Cash/Bank $2,250
Cr Account Receivable $2,250
Step-by-step explanation:
At the time the transaction was carried out, that is the previous month, the accounting entries was a debit to account receivable (client account) and a credit to sales account. which is as shown below:
<em>Dr Account Receivable $2,250</em>
<em>Cr Sales $2,250</em>
<em>Being sales on credit to a client.</em>
The client account will have an outstanding debit balance of <em>$2,250 </em>until payment is made. once payment is made, the following entries will be made to ensure no outstanding balance in the client account.
Dr Cash/Bank $2,250
Cr Account Receivable $2,250.
To find missing lengths of sides
hope this helped have a nice day