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:
1 Box Weighed 180 Pounds, Because 2160 Divided By 12 Is 180. Hope That Helps
Answer:
(c) BC ≅ BC, reflexive property
Step-by-step explanation:
The conclusion of this proof derives from CPCTC and the SAS congruence postulate. In order for SAS to apply, corresponding sides and the angle between them must be shown to be congruent. The congruence statement ...
ΔABC ≅ ΔDCB
tells you these pairs of sides and angles are congruent:
- AB ≅ DC . . . . statement 2
- ∠ABC ≅ ∠DCB . . . . statement 4
- BC ≅ CB . . . . (missing statement 5)
- AC ≅ DB . . . . statement 7
That is, the statement needed to complete the proof is a statement that segment BC is congruent to itself. That congruence is a result of the reflexive property of congruence.
Answer:
Not sure about part B but A is 13 and i'm not sure what you mean by round it to the nearest hundredth place value wouldn't that be 300
Step-by-step explanation:
5212g is the correct answer
