Answer:
x > -2.5
Step-by-step explanation:
1) 5.1 (3 + 2.2x) > -14.25 - 6 (1.7x + 4)
2) 15.3 + 11.22x > -14.25 - 10.2x - 24
3) 15.3 + 11.22x > -38.25 - 10.2x
4) 11.22x + 10.2x > -38.25 - 15.3
5) 21.42x > -53.55
6) x > -2.5
Alternate forms
x > -5/2x or x > -2 1/2
Answer:
Step-by-step explanation:
First and foremost, all quadratics have a domain of all real numbers (as long as we are not given only a portion of the graph, or one with endpoints. Our graph does not have endpoints, so it is assumed that the tails will continue to go down into negative infinity and at the same time, the x coordinates will keep growing as well.) Since our quadratic is upside down, it has a max. That means that none of the values on the graph will be above that point. All the values will be below that highest point (the highest y-value). Y-values indicate range, and since our highest y-value is at y = 2, then the range is
y ≤ 2
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:
Answer:
i think they are y and (x,y)
