The derivative of 1/logx is With the chain rule.
1log(x)=log(x)−1 is ,= -1xlog(x)2 .
The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".
The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

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9514 1404 393
Answer:
log(128) = t·log(2)
Step-by-step explanation:
You can take logarithms the way the formula is, or you can simplify it first.
<u>As is</u>
log(10000) = log(625) -log(8) +t·log(2)
__
<u>Simplified</u>
8·10000/625 = 2^t . . . . . multiply by 8/625
128 = 2^t
log(128) = t·log(2)
_____
Solution:
t = log(128)/log(2) = 7
The external angle (53°) is half the difference of the intercepted arcs (y°, 180°), so we have
53 = (180 - y)/2
106 = 180 - y
y = 180 - 106
y = 74
The appropriate choice for the value of y is ...
74
The matrix that represents the matrix D is ![\left[\begin{array}{cccc}3&1&-9&8\\2&2&0&5\\16&1&-3&11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%261%26-9%268%5C%5C2%262%260%265%5C%5C16%261%26-3%2611%5Cend%7Barray%7D%5Cright%5D)
<h3>How to determine the matrix d?</h3>
Given the elements of the matrix C.
The matrix c is represented by its rows and columns element, and the arrangements are:
C11 = 3 C12 = 1 C13=-9 C14 = 8
C21 = 2 C22=2 C23 =0 C24 = 5
C31 = 16 C32 = 1 C33=-3 C34=11
Remove the matrix name and position
3 1 9 8
2 2 0 5
16 1 -3 11
Represent properly as a matrix:
![C = \left[\begin{array}{cccc}3&1&-9&8\\2&2&0&5\\16&1&-3&11\end{array}\right]](https://tex.z-dn.net/?f=C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%261%26-9%268%5C%5C2%262%260%265%5C%5C16%261%26-3%2611%5Cend%7Barray%7D%5Cright%5D)
Matrix C equals matrix D.
So, we have:
![D = \left[\begin{array}{cccc}3&1&-9&8\\2&2&0&5\\16&1&-3&11\end{array}\right]](https://tex.z-dn.net/?f=D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%261%26-9%268%5C%5C2%262%260%265%5C%5C16%261%26-3%2611%5Cend%7Barray%7D%5Cright%5D)
Hence, the matrix that represents matrix D is ![\left[\begin{array}{cccc}3&1&-9&8\\2&2&0&5\\16&1&-3&11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%261%26-9%268%5C%5C2%262%260%265%5C%5C16%261%26-3%2611%5Cend%7Barray%7D%5Cright%5D)
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W = 120 + 5(h - 30) <== ur equation....where w = total wages earned and h = number of hrs worked
** I got the 120 from (30 hrs * $ 4 per hr) = $ 120