1.
The base can't be negative, therefore 
.
2.
3.
Answer: y ≥ (3/5)*x - 3
Step-by-step explanation:
In the graph, we can see that we are above a bold line, that goes through the points (0, -3) and (5, 0)
First, let's find the equation for this line:
y = a*x + b
the value of a is the slope and is equal to:
a = (0 - (-3))/(5 -0) = 3/5
and the value of b is the point where the line intersects the y-axis, in this case, b = -3
then our line is
y = (3/5)*x - 3
As the shaded part is above the line, this equality represents the minimum value that y can take for a given x, and because the line is not a doted line, we know that the equality is valid, so we must use the ≥ symbol.
y ≥ (3/5)*x - 3
Answer:
24,12,6
Step-by-step explanation:
48 is even, so you can factor out 2, your first prime factor
<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
