X should be 7 after doing all the calculation.
The answer is 25 I think sorry if it’s wrong :)
Beth's description of the transformation is incorrect
<h3>Complete question</h3>
Beth says that the graph of g(x)=x-5+1 is a translation of 5 units to the left and 1 unit up of f(x) = x. She continues to explain that the point (0,0) on the square root function would be translated to the point (-5,1) on the graph of g(x). Is Beth's description of the transformation correct? Explain
<h3>How to determine the true statement?</h3>
The functions are given as:
g(x) = x - 5 + 1
f(x) = x
When the function f(x) is translated 5 units left, we have:
f(x + 5) = x + 5
When the above function is translated 1 unit up, we have:
f(x + 5) + 1 = x + 5 + 1
This means that the actual equation of g(x) should be
g(x) = x + 5 + 1
And not g(x) = x - 5 + 1
By comparison;
g(x) = x - 5 + 1 and g(x) = x + 5 + 1 are not the same
Hence, Beth's description of the transformation is incorrect
Read more about transformation at:
brainly.com/question/17121698
#SPJ1
By definition of the derivative,
![\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B%28s%20%2B%20h%29%5E3%7D2%20%2B%201%5Cright%29%20-%20%5Cleft%28%5Cfrac%7Bs%5E3%7D2%20%2B%201%5Cright%29%7D%7Bh%7D)
![\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7Bs%5E3%2B3s%5E2h%2B3sh%5E2%2Bh%5E3%7D2%20%2B%201%5Cright%29%20-%20%5Cleft%28%5Cfrac%7Bs%5E3%7D2%20%2B%201%5Cright%29%7D%7Bh%7D)
![\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac%7B%5Cfrac%7B3s%5E2h%2B3sh%5E2%2Bh%5E3%7D2%7D%7Bh%7D)
![\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac12%20%5Cfrac%7B3s%5E2h%2B3sh%5E2%2Bh%5E3%7D%7Bh%7D)
![\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac12%20%283s%5E2%2B3sh%2Bh%5E2%29)
![\displaystyle\frac{dr}{ds} = \frac{3s^2}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdr%7D%7Bds%7D%20%3D%20%5Cfrac%7B3s%5E2%7D2)