Double is double, so you can solve
.. 2 = 1*e^(.08t)
and get the correct value of t.
Taking logs,
.. ln(2) = .08t
.. ln(2)/.08 = t ≈ 8.7 . . . . years
Answer:
3/4 using equation y2-y1/x2-x1
Step-by-step explanation:
Use the equation y2-y1/x2-x1
x1, y1 x2, y2
(16,12) (0,0)
0-12/0-16= -12/-16= 3/4
Hello!
The greatest common factor (GCF) is self explanatory. We find the factors of each number, and find the largest ones that are in common
12: 1,12,2,6,3,4
33:1,33,3,11,
As you can see, the greatest number these two have in common is 3.
Now for the next set.
45: 1,45,3,15,5,9
70:1,70,2,35,5,14,7,10
As you can see, our GCF is 5.
Therefore, our answers are below.
9) 3
10) 5
I hope this helps!
Answer:
The values of x for which the model is 0 ≤ x ≤ 3
Step-by-step explanation:
The given function for the volume of the shipping box is given as follows;
V = 2·x³ - 19·x² + 39·x
The function will make sense when V ≥ 0, which is given as follows
When V = 0, x = 0
Which gives;
0 = 2·x³ - 19·x² + 39·x
0 = 2·x² - 19·x + 39
0 = x² - 9.5·x + 19.5
From an hint obtained by plotting the function, we have;
0 = (x - 3)·(x - 6.5)
We check for the local maximum as follows;
dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0
6·x² - 38·x + 39 = 0
x² - 19/3·x + 6.5 = 0
x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2
∴ x = 1.288, or 5.045
At x = 1.288, we have;
V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99
V ≈ 22.99 in.³
When x = 5.045, we have;
V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023
Therefore;
V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5
The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.
51x0.15=7.65 7.65 is the 15% of 51
