Hello,
any point equidistant from the ends of a segment belongs to the perpendicular bisector of the segment.
|AD|=|BD| and |AC|=|BC|
For this, it is kind of like a triangle and having to find the diagonal. Let's see what the two legs of the triangle would be.
Leg one= 7 units
Leg two= 8 units
8²+7²= c²
64 + 49=113
√113≈10.6
The line is 10.6 units long.
It was reduced because as you can see on the scale the actual picture has larger spaces between the numbers.
2 is the correct answer I'm almost definite
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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