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

53°
Step-by-step:
- All angles in a triangle add up to 180°.
- Since your triangle is a right angle triangle, one of the angles are 90°.
- You already know another one, 37°.
- All you have to do to find the last one is to subtract the two you already know from 180.
180 - (90+37)
= 180 - 127
= 53
The measure of the last angle is 53°.
Answer:
10 6/10
Step-by-step explanation:
8/6 = 1.333
1.333 x 8 = 10.667
as a mixed number: 10 6/10
Answer:
Step-by-step explanation:
50% = 0.5
(5 * 5) + (0.5 * 2) =
25 + 1 =
26 <===
2•3=6
I’m not for sure but used photomath