The diameter of Circle Q terminates on the circumference of the circle at (0,3) and (0,-4). Write the equation of the circle in
1 answer:
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Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>Step-by-step explanation:
Given the function j(x) = 11.6 and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6 and k(x) =
are inverse functions.
Answer:
<h2>{5}</h2>Step-by-step explanation:
Answer:
The Point C shows the location of 5-2i in the complex plane: 5 points to the right of the origin and 2 points down from the origin.
Step-by-step explanation:
We have the complex number 5-2i and we have to show the location of the point that represents that number in the complex plane
In the complex plane the real numbers are located in the horizontal axis, increasing to the right. The positives real numbers are at the right of the origin and the negatives to the left.
The complex numbers are located in the vertical axis, with the positives over the origin and the negatives below the origin.
This complex number 5-2i is the sum of a real part (5) and a imaginary part (-2i), so the point will be 5 units rigth on the horizontal axis (for the real part) and 2 units down in the vertical axis (for the imaginary part).
