It's a doubling time question, so the formula will have the form of y = a(2)^(x/t). Y is the present amount; a is the initial amount; x is the number of year; t is the period of doubling time.
So now you just plug in values:
y = 5000(2)^ (36/12)
y = 5000(2^3)
y = 5000 (8)
y = 40 000
And it's the same for the second part (just with 48 instead of 36):
y = 5000(2)^ (48/12)
y = 5000(2^4)
y = 5000 (16)
y = 80 000
The answer is 3b. hope it helps
1)
C represents more of a linear model for the data because it has equal amount of points on both sides of the line.
2)
This equation is set up in y=mx+b
M is your slope and b is your y-intercept.
Your slope is -2. Y-intercept is 24.
y=-2x+24
3)
When J=2, the line his K at 20.
4)
This equation is set up in y=mx+b
M is your slope and b is your y-intercept.
Your slope is 15. Y-intercept is 40.
y=15x+40
Let's first rewrite it in vertex form.
Start by completing the square.
y = x² + 4x + 4 - 4 - 1
y = (x + 2)² - 5
(x + 2)² = y + 5
Now, 4a = 1; a =

So, the vertex form is (x + 2)² = 4(

)(y + 5)
So, we know that the vertex is at (-2, -5).
Since it's a concave up parabola, we just have to change the y-value to find the focus.
∴ F(-2, -5 +

)
F(-2,

)