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The question is incomplete, as the depth of the hook isn't given in the question. However, using a proposed depth for the hook, we could give a stepwise solution to the exercise.
Answer:
Kindly check explanation
Step-by-step explanation:
If given a hook depth of 6 meters, and a distance of 8 m(fishing line from hook) at the the Same depth, then the distance between rose and the fish, x can be calculated, using the relation.
Hypotenus² = opposite² + adjacent²
(refer to attached picture)
x² = 6² + 8²
x² = 36 +. 64
x² = 100
x = sqrt(100)
x = 10 meters
You can use the actual depth value to obtain the actual solution to your question.
There's more than one way to combine them really
but an obvious one will be
![\bf \begin{array}{llll} h(x)&=&(f\circ g)(x)\\\\ &&\sqrt[3]{7x+1}\\\\ &&\sqrt[3]{g(x)}\leftarrow \begin{array}{llll} f(x)=\sqrt[3]{x }\\\\ g(x)=7x+1 \end{array} \end{array}\\\\ -----------------------------\\\\ (f\circ g)(x)\iff f[\quad g(x)\quad ]=\sqrt[3]{g(x)}\implies f[\quad g(x)\quad ]=\sqrt[3]{7x+1}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0Ah%28x%29%26%3D%26%28f%5Ccirc%20g%29%28x%29%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7B7x%2B1%7D%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cleftarrow%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7Bx%20%7D%5C%5C%5C%5C%0Ag%28x%29%3D7x%2B1%0A%5Cend%7Barray%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28f%5Ccirc%20g%29%28x%29%5Ciff%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cimplies%20%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7B7x%2B1%7D)
but an obvious one will be