Given
![x+1 = \sqrt{7x+15}](https://tex.z-dn.net/?f=x%2B1%20%3D%20%5Csqrt%7B7x%2B15%7D)
We have to set the restraint
![x+1\geq 0 \iff x \geq -1](https://tex.z-dn.net/?f=x%2B1%5Cgeq%200%20%5Ciff%20x%20%5Cgeq%20-1)
because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:
![(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0](https://tex.z-dn.net/?f=%28x%2B1%29%5E2%3D7x%2B15%20%5Ciff%20x%5E2%2B2x%2B1%3D7x%2B15%20%5Ciff%20x%5E2-5x-14%20%3D%200)
The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.
Similarly, we have
![x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}](https://tex.z-dn.net/?f=x-3%20%3D%20%5Csqrt%7Bx-1%7D%2B4%20%5Ciff%20x-7%3D%5Csqrt%7Bx-1%7D)
So, we have to impose
![x-7\geq 0 \iff x \geq 7](https://tex.z-dn.net/?f=x-7%5Cgeq%200%20%5Ciff%20x%20%5Cgeq%207)
Squaring both sides, we have
![(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0](https://tex.z-dn.net/?f=%28x-7%29%5E2%3Dx-1%20%5Ciff%20x%5E2-14x%2B49%3Dx-1%20%5Ciff%20x%5E2-15x%2B50%20%3D%200)
The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.
Step-by-step explanation:
here is the answer to your question
Answer:
16 km
Step-by-step explanation:
the radius is half of the diameter