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
Least common mulitplue is the smallest number that has 49 and 28 as factors
so to do this
find the factor so 49 and 28 and gropu all of them so both are in it
49=7 times 7
28=2 times 2 times 7
therefor yo need at least two 7's and two 2's
that equals
7 times 7 times 2 itme s2=49 times 4=196
answer is 196
What is the question? no question has been asked. need more info to solve.
Answer:
4 people
Step-by-step explanation:
Let's first write an equation.
We want to find the number of tickets t
The total was $49 so our equation will equal $49 and they spent $15 in snacks so on the other side of the eqaution we have to add $15, on the same side as the snacks we need to add in the cost of tickets, which is $8.50 multiplies by t, or the total number of tickets.
$49 = $8.50t + $15
Solve for t, first subtract $15 from both sides.
$49 - $15 = $8.50t
$34 = $8.50t Divide both sides by $8.50
$34/$8.50 = $8.50t/$8.50
$34/$8.50 = t
4 = t
4 tickets meaning 4 people went.
Answer Standard form
Step-by-step explanation:
