Recall that variation of parameters is used to solve second-order ODEs of the form
<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>
so the first thing you need to do is divide both sides of your equation by <em>t</em> :
<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>
<em />
You're looking for a solution of the form

where


and <em>W</em> denotes the Wronskian determinant.
Compute the Wronskian:

Then


The general solution to the ODE is

which simplifies somewhat to

-5.2 -4.8= -10 thats your answer and it easy math.
Let
A=8t-3
B=6t+5
we know that
<span>total earnings=A+B
</span>so
total earnings=(8t-3)+(6t+5)------> (8t+6t)+(-3+5)------> (14t+2) <span>dollars
the answer is
</span>(14t+2) dollars<span>
</span>
you would just add both of them together so the weight would be 7.69 Ibs
Answer:
-4
Step-by-step explanation:
If you convert the words into an equation, you get:
-6+2
-6+2=-4
-4 is your answer.
<em>~Stay golden~ :)</em>