<h3>
Answers:</h3>

================================================
Work Shown:
Part 1

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.
------------------
Part 2

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2
In this case, a = sqrt(x-1) and b = 5.
You could also use the box method as a visual way to expand out 
Answer:
D
Step-by-step explanation:
Answer:
Rui will reach his lowest altitude after diving for 10 seconds
Step-by-step explanation:
In the quadratic equation y = ax² + bx + c, the coordinates of the lowest or the highest point is (h, k), where
- h =
and k = y at x = h
In the given question
∵ d(x) =
x² - 10 x
∵ d represents his altitude in meters after x seconds
→ Compare it with the form above to find a and b
∴ a =
and b = -10
∵ h =
→ Substitute the values of a and b in the rule to find h
∴ h =
= 10
∴ h = 10
∵ h is the x-coordinate of the lowest point
∴ The time of the lowest altitude is 10 seconds
∴ Rui will reach his lowest altitude after diving for 10 seconds