Answer:
1537500
Step-by-step explanation:
it should be that. all you do is multiply 615000 by 2.5
The pattern of the trend line from what we can see here shows the existence of a positive upward trend.
<h3>What is a positive upward trend?</h3>
We can see that the graph has fluctuations which have an upward trend over the years. This upward trend can be gotten from the straight line that was drawn on the graph.
It slants to the top. This shows us that the trend is positive and rising over the period of time that we have in the graph. In the graph, the upward trend can be seen from the fact that there has been a rise in the temperature of the city over the time period that was illustrated. The increase in temperature rose from 31.5 to 33 degrees period of time.
Hence we can conclude that The pattern of the trend line from what we can see here shows the existence of a positive upward trend.
Read more on trend lines here: brainly.com/question/27194207
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Answer:
x= -3/5
Step-by-step explanation:
4x-2y=14
y=12x-1
4x-2(12x-1)=14
4x-24x+2=14
4x-24x=12
-20x=12
×= -12/20
x= -3/5
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
