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>
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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

Are you tryna solve for x?
Answer:
A. False
B. True
C. True
D.True
Step-by-step explanation:
A. False . The significance level or alpha is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference. 0.01 alpha is better than 0.05 alpha . 0.01 indicate a 1% risk of rejecting the null hypothesis when it is true .
B. True . If the p-value is less than alpha, we reject the null hypothesis . Therefore statistically significant.
C . True . If the p-value is less than alpha, we reject the null hypothesis
D. True . Alpha will be greater than p-value . Therefore we will reject .
50 per share hope this helped
STEP 1:
Divide the length of the ribbon by the size of the pieces.
12 inches= 1 foot
= 9 ft ÷ 15 inch pieces
convert ft to inches
= (12)(9) ÷ 15
= 108 ÷ 15
= 7.2 pieces from one roll
STEP 2:
multiply the number of pieces that can be cut from one roll by the number of rolls
= 7.2 pieces * 5 rolls
= 36 pieces total from 5 rolls
ANSWER: The greatest number of 15-inch pieces that can be cut from five 9-foot rolls is 36.
Hope this helps! :)