H=-16x to the second power +136
Solution:
Given: zj = 3- j2, zz = - 4 + j3
Part a:
zj = 3- j2 = 
zz = - 4 + j3 = 
Part b:
|zj| = |3− j2| = √3² +(−2)² = √13 = 3.60
|z1| = √(3− j2)(3+ j2) = √
13 = 3.60.
Part c:
With the help of part a:
z1z2 = 
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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Answer:
x = 12
Step-by-step explanation:
3х+7 = 5x-17
-2x = -24