1. 95-87 = 8 /2 = 4
2. 91 - 82 = 9 /2 = 4.5 =5
3. 91 - 87 = 4
4. B 1 times
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.
To solve this problem you must apply the proccedure shown below:
1. You have to r<span>ewrite x=12 in polar form. Then, you have:
12=rCos</span><span>θ
2. Then, you must solve for r, as following:
r=12/Cos</span><span>θ
</span> 3. You have that 1/Cosθ=Sec<span>θ, therefore:
</span> r=12(1/Cos<span>θ)
</span> r=12Sec<span>θ
</span> Therefore, as you can see, the answer is: r=12Secθ<span>
</span>
317 is the answer you are looking for.