Answer:
7/44
Step-by-step explanation:
you can't simply subtract 1/11 from 1/4 because the denominators are not the same . Meaning you have to convert the denominators into a similar number. Transformers in even number and a consecutive number while 11 is an odd number and a prime number they don't really agree on anything 11 can only be divided by itself and 1 wall for can be divided by a multitude of things. Because of them not exactly agreeing on any specific category , you have to multiply them by each other . So your new fractions should look like 11 / 44 and 4 / 44 . from there you can easily subtract 4 from 11 and get 7 / 44 now normally you can reduce these types of fractions but because seven can only be divided by itself and 44 is not a factor of 7 you cannot reduce this fraction .
Answer:
54 -3n²
Step-by-step explanation:
The square of a number (n) is represented by n². Three times that value is represented by 3n².
The relation "a is subtracted from b" is represented as ...
b - a
When 3n² is subtracted from 54, the appropriate representation is ...
54 -3n²
Answer:
6x + 2(−2x + 1) = 22
Step-by-step explanation:
6x + 2(−2x + 1) = 22
This step can be used to solve by substitution method.
Answer:
28 tickets
Step-by-step explanation:
252/9=28
Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:
a) x’=t–sin(t), x(0)=1
Apply integral both sides:
where k is a constant due to integration. With x(0)=1, substitute:
Finally:
b) x’+2x=4; x(0)=5
Completing the integral:
Solving the operator:
Using algebra, it becomes explicit:
With x(0)=5, substitute:
Finally:
c) x’’+4x=0; x(0)=0; x’(0)=1
Let 
be the solution for the equation, then:
Substituting these equations in <em>c)</em>
This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:
Finally:
