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saw5 [17]
3 years ago
9

Match the information on the left with the appropriate equation on the right.

Mathematics
1 answer:
Eduardwww [97]3 years ago
8 0
3)7?,79 that the same name of this game I
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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

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}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

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}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Solve the equation 42=7r r=​
Juli2301 [7.4K]

Answer: r=6

Step-by-step explanation:

Because if 42 equals 7 times r you just have to divide 42 by 7.

8 0
3 years ago
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Let v1=(1, 6) and v2=(5, -3)
astraxan [27]

Answer:

See Explanation;

Step-by-step explanation:

The given vectors are;

v_1=\:\: and v_2=\:\:

a) To add or subtract two vectors, we add or subtract their corresponding components

v_1+v_2=\:\:+\:\:

v_1+v_2=\:\:

v_1+v_2=\:\:

v_1-v_2=\:\:-\:\:

v_1-v_2=\:\:

v_1-v_2=\:\:

v_2-v_1=\:\:-\:\:

v_2-v_1=\:\:

v_2-v_1=\:\:

b) See attachment for graphs(1,6

3 0
3 years ago
A total of 480 were sold for the school play. They were either adult tickets or students tickets. The number of student tickets
Talja [164]

Ok so basically, the number of student tickets is 3x, where x=the number of adult tickets sold. And we know that s(for student tickets)+x=480 total tickets sold. So if we replace s with 3x we have 3x+x=480, or 4x=480. We divide by 4 and get x=120, which is the amount of adult tickets sold.

5 0
3 years ago
What is 128.79$ with 30% discount
g100num [7]
ANSWER:
128.79 x .30 = $38.64
7 0
3 years ago
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