Look for an integrating factor 
:
Multiply both sides by 
:
Condense the left side as the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%28t%2B1%29%5E%7B-7%2F9%7Dy%5Cright%5D%3D%5Cdfrac%7B14%7D9t%28t%2B1%29%5E%7B-16%2F9%7D)
Integrate both sides:
For the integral on the right, substitute
Given that 
, we get
Answer:
8
Step-by-step explanation:
First you subtract 20 by 4. Then you get 16. 16 divided by 2 gets you 8. 8 is the answer.
Hope this helps dude
Answer:
No
Step-by-step explanation:
5-2=3
Inverse/reciprocal =7
Answer:
The answer is D
Step-by-step explanation:
4+(-8+24)=20
4+(16)=20
20=20
Answer: x = "-1 " ; <u>and</u>: "(1 ± √3) / 2" .
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Explanation:
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Given: 3x³ − 2x + 1 = 0 ; Find "x" ;
Factor: "3x³ − 2x + 1" ;
3x³ − 2x + 1 = (3x² <span>− 3x + 1)(x + 1) ;
Now, set the equation equal to "0" (zero);
</span> (3x² − 3x + 1)(x + 1) = 0 ;
x = 0 ; when:
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(x + 1) = 0 ; and/or: when:
(3x² − 3x + 1) = 0 ;
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Start with "x + 1 = 0" ;
Subtract "1" from each side of the equation ; to isolate "x" on one side of the <span>equation ; and</span> to solve for "x" ;
x + 1 − 1 = 0 − 1 ;
x = -1 ;
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Now, try:
3x² − 3x + 1 = 0 ;
Since: "3x² − 3x + 1 " cannot be factored; and since:
"3x² − 3x + 1 = 0" ; is written in the "quadratic equation format"; that is:
"ax² + bx + c = 0 ; a ≠ 0; We can use the "quadratic equation formula" to solve for "x" ;
in which "a = 3" ;
"b = -3" ;
"c = 1" ;
The quadratic equation formula is:
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x = [-b ± √(b²−4ac] / 2a ;
= ? (Let us plug in our known values for "a, b, & c"l
x = { [-(-3] ± [√[(-3²) − (4*3*1)] } / {2* 3} ;
x = [3 ± √(9 − 12)] / 6
x = [3 ± √(-3 )] / 6 ;
x = (1 ± √3) / 2 .
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So, the answers are: x = " -1 " ; and: "</span>(1 ± √3) / 2 " .
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