Express the following as:ax+by+c=0-2x+3y=6
2 answers:
The expression -22x + 3y = 6 in the form ax + by + c = 0 is -2x + 3y - 6 = 0
<h3>Linear Equations</h3>
The given equation is:
-2x + 3y = 6
Comparing the equation above with ax + by + c = 0
Rewrite -2x + 3y = 6 in the form ax + by + c = 0
-2x + 3y - 6 = 0
If -2x + 3y - 6 = 0 is compared with ax + by + c = 0
a = -2, b = 3, c = -6
Therefore, the expression-22x + 3y = 6 in the form ax + by + c = 0 is -2x + 3y - 6 = 0
Learn more on linear equations here: brainly.com/question/14323743

- <u>it </u><u>can </u><u>be </u><u>expressed </u><u>as </u><u>-2x </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>6</u>
<u>on </u><u>comparing </u><u>with </u><u>ax+by+c=0</u>
<u>We </u><u>can </u><u>find </u><u>out </u><u>that </u>



<h2><u>hope</u><u> it</u><u> helps</u></h2>
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Answer: The required derivative is 
Step-by-step explanation:
Since we have given that
![y=\ln[x(2x+3)^2]](https://tex.z-dn.net/?f=y%3D%5Cln%5Bx%282x%2B3%29%5E2%5D)
Differentiating log function w.r.t. x, we get that
![\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5Bx%27%282x%2B3%29%5E2%2B%282x%2B3%29%5E2%27x%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5B%282x%2B3%29%5E2%2B2x%282x%2B3%29%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B4x%5E2%2B9%2B12x%2B4x%5E2%2B6x%7D%7Bx%282x%2B3%29%5E2%7D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B8x%5E2%2B18x%2B9%7D%7Bx%282x%2B3%29%5E2%7D)
Hence, the required derivative is 