Answer:
I'm sure the answer to this is: x<20/3
If you would like to write a * b + c in simplest form, you can do this using the following steps:
a = x + 1
b = x^2 + 2x - 1
c = 2x
a * b + c = (x + 1) * (x^2 + 2x - 1) + 2x = x^3 + 2x^2 - x + x^2 + 2x - 1 + 2x = x^3 + 3x^2 + 3x - 1
The correct result would be x^3 + 3x^2 + 3x - 1.
Answer:
12 devided by 25 then multiply by 100
48%
Answer:
- Initial amount of the material is 67 kg
- Hal-life is 83 hours
<u>The required equation is:</u>
- m(x) = 67 *
, where m- remaining amount of the radioactive material, x - number of hours
<u>After 5 hours the material remains:</u>
- m(5) = 67 *
= 64.260 (rounded)
The correct question is:
Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)
Answer:
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
Step-by-step explanation:
We need to verify that
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
We differentiate
x = c1e^(-t) + c2e^(3t)
twice in succession, and substitute the values of x, x', and x'' into the differential equation
x''- 2x' - 3x = 0
and see if it is satisfied.
Let us do that.
x = c1e^(-t) + c2e^(3t)
x' = -c1e^(-t) + 3c2e^(3t)
x'' = c1e^(-t) + 9c2e^(3t)
Now,
x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]
= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)
= 0
Therefore, the differential equation is satisfied, and hence, x is a solution.
