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
![5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.](https://tex.z-dn.net/?f=5m%5E2e%5E%7Bmt%7D%2B3me%5E%7Bmt%7D-2e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%285m%5E2%2B3y-2%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B3m-2%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B5m-2m-2%3D0%5C%5C%5C%5C%5CRightarrow%205m%28m%2B1%29-2%28m%2B1%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B1%29%285m-1%29%3D0%5C%5C%5C%5C%5CRightarrow%20m%2B1%3D0%2C~~~~~5m-1%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-1%2C~%5Cdfrac%7B1%7D%7B5%7D.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
![y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D2.8%5C%5C%5C%5C%5CRightarrow%20-A%2B%5Cdfrac%7BB%7D%7B5%7D%3D2.8%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D14%5C%5C%5C%5C%5CRightarrow%20-5A-A%3D14~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7BUisng%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-6A%3D14%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B14%7D%7B6%7D%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B7%7D%7B3%7D.)
From equation (ii), we get
Thus, the required solution is
C) 1059 skittles
Step-by-step explanation:
We know that the first container held 192 skittles. Knowing the dimension of the container we can calculate the volume of 192 skittles:
Volume of 192 skittles = 5 × 4 × 4 = 80 cm³
Volume of 1 skittle = 80 / 192 = 0.42 cm³
We also know that the second container held 258 skittles. Knowing the dimension of the container we can calculate the volume of 258 skittles:
Volume of 258 skittles = 12 × 3 × 3 = 108 cm³
Volume of 1 skittle = 108 / 258 = 0.42 cm³
We found that the volume of 1 skittle is equal to 0.42 cm³. Now we calculate the volume of the skittles jar:
volume of cylinder = π × radius² × height
volume of skittles jar = 3.14 × 3.5² × 11.5 = 442 cm³
Now we can calculate the number of skittles in the jar:
number of skittles in the jar = volume of the jar / skittle volume
number of skittles in the jar = 442 / 0.42 = 1052 which is close the C) 1059
Learn more about:
volume of cylinder
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Answer:
34.576 = 30 + 4 + 0.5 + 0.07 + 0.006
Answer:
x=50
y=50
Step-by-step explanation:
its right because sas
Answer: £210
Step-by-step explanation:
you do the cm they given you so if it’s 630cm x 120cm = 75,600
30cm x 30cm = 900cm
75,600 divided by 900 = 84
£2.50 x 84 = £210
