Answer:
The percent of the area under the density curve where 
is more that 3 is 25 %.
Step-by-step explanation:
Since the density curve is a linear function, the area under the curve can be calculated by the geometric formula for a triangle, defined by the following expression:
(1)
Where:
- Area, in square units.
- Base of the triangle, in units.
- Height of the triangle, in units.
The percent of the area is the ratio of triangle areas under the density curve multiplied by 100 per cent, that is:
The percent of the area under the density curve where is more that 3 is 25 %.
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
The unit rate will be in revolutions per second. That means divide number of revolutions by number of seconds.
first convert from mixed fraction.
6 2/5 = 32/5 revolutions
2 2/3 = 8/3 seconds
32/5 ÷ 8/3
flip multiply
32/5 × 3/8 = 12/5 revolutions per second
12/5 = 2 2/5 revolutions per second
Answer and Step-by-step explanation:
Given Asin(wt + phi), we know that sin (A + B) = sinAcosB + sinBcosA. This means:
Asin(wt + phi) = Asin(wt)cos(phi) + Asin(phi)cos(wt).
Let Acos(phi) = c2 and Asin(phi) = c1 we have:
Asin(wt + phi) = c2sin(wt) + c1cos(wt)
