Answer:
After 2 hours: -6°C.
After 5 hours: -15°C.
After hours: -1.5°C
1 hour ago: 3°C.
3 hours ago: 9°C.
4.5 hours ago: 13.5°C.
To find these values, you simply need to multiply the hours elapsed by 3°C. Subtract from 0°C for future temperatures, and ADD to 0°C for the past.
Let's say you are solving for the temperature in 5 hours. All you need to do is:
3°C× 5= 15°C. 0°C-15°C= -15°C
For the temperature 3 hours AGO, what you will need to do instead is:
3°C×3= 9°C. 0°C+ 9°C= 9°C.
Answer:
f(x) = (x -2)(x -1+3i)(x -1-3i)
Step-by-step explanation:
You can use synthetic division to find the remaining quadratic factor in the cubic. Then any of the usual means of solving the quadratic will help you find its linear factors.
In the attached, I show the synthetic division, the factoring to real numbers, and the solution that finds the complex linear factors by completing the square.
Of course, you know that for zeros a, b, and c, the linear factors are ...
f(x) = (x -a)(x -b)(x -c)
Here, we have a=2, b=1-3i, c=1+3i.
f(x) = (x -2)(x -1+3i)(x -1-3i)
