Answer:
OPTION C: h(x) = x - 3
Step-by-step explanation:
To find the inverse of the function y = x + 3.
Swap 'x' and 'y'.
We get x = y + 3.
Solve for 'y' now.
We get y = x - 3
Therefore, the inverse function becomes h(x) = x - 3.
Answer:
The numbers are 12 and 3.
Step-by-step explanation:
We can solve this problem by working with the information we have and setting up some equations.
We know that one number is four times as large as another. So, let the smaller number be represented by the variable x and the bigger number be represented by 4x, since it is four times as large.
Now, we know that if the numbers are added together, then the result is six less than seven times the smaller number. This can also be represented by the equation 4x + x = 7x - 6.
Let's solve that equation like so:
So, the smaller number must be 3 (remember that x represented the smaller number). To find the bigger number, all we need to do is multiply 3 by 4, which gives us 12. Therefore, the numbers are 12 and 3.
9514 1404 393
Answer:
55,637.8 square inches
Step-by-step explanation:
We can find side n using the Law of Sines:
n/sin(N) = p/sin(P)
n = p(sin(N)/sin(P)) = 600·sin(64°)/sin(96°)
n ≈ 542.246913 . . . . inches
The angle O is ...
O = 180° -N -P = 180° -64° -96° = 20°
Then the area is ...
A = 1/2·np·sin(O)
A = (1/2)(542.246913 in)(600 in)·sin(20°) ≈ 55,637.81008 in²
The area of ∆NOP is about 55,637.8 in².
Problems of this sort are frequently found in physics. If you study calculus or physics you'll learn how to create the equation representing the velocity of an object in flight.
Here, you don't need to calculate velocity, but rather time. Start with this equation:
v = v0 + a t^2, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration due to gravity (denoted by g instead of a), and t is the elapsed time.
You are told that v0 is 15 ft/sec. Set v = to 0, as the ball stops moving for the tiniest instant at the top of its trajectory. Use g = - 32 ft (per second squared).
Then 0 = 15 ft/sec - 32 [ft/(seconds squared)] t.
Solve this for t. This is the time required for the ball to come to a complete stop at the top of its trajectory.
Finally, multiply this time by 2, since the ball begins to fall and returns to its original height.