Answer:
A, D, E
Step-by-step explanation:
Sorry for being late
Answer:
1) f(g(-2))=-3, 2) g(f(0))=5
Step-by-step explanation:
1)f(g(-2))
First, g(-2) means that x=-2. Hence, you must find the value of g(x) in the table when x=-2. You can see that, when x=-2, g(-2) = -3.
Next, you must find f(-3) in the graph, where x=-3. You can see in the graph that, when x=-3, f(-3) = -2.
Therefore, f(g(-2))=-3
2) g(f(0))
In the case, we must apply the inverse procedure. First, check in the graph that, when x=0, f(0) =1.
Next, we must look at the table and see that, when x=1, g(1)=5. Hence,
g(f(0))=5
Step-by-step explanation:
Jupiter's
Jupiter's menagerie of moons includes the largest in the solar system (Ganymede), an ocean moon (Europa) and a volcanic moon (Io). Many of Jupiter's outer moons have highly elliptical orbits and orbit backwards (opposite to the spin of the planet).
i want 3 brainliest
Now,
(2x + y) (3x - 7)
2x (3x - 7y) + y (3x - 7)
6x^2 - 14xy + 3xy - 7y
6x^2 - 11xy - 7y
This is Right answer...

In the given figure, NQ acts as a diameter and since diameter subtends 90° at the arc of the circle, so we can conclude that ~



Now, Let's use Angle sum property of Triangle to solve for z :







I hope you understood the procedure, you can ask me in comments if you have any doubts.