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Vinvika [58]
2 years ago
13

In your notebook, set up the following subtraction in a vertical format and select the correct answer. Find 2p 2 3p - 4 less - 2

p 2 - 3p 4. 4p2 6p 8 4p2 - 6p - 8 4p2 6p - 8 0.
Mathematics
1 answer:
Elena L [17]2 years ago
3 0

The equation in the reduced form of the equation is \rm 4p^ 2 + 6p - 8.

Given that,

Equation; \rm  2p^ 2 + 3p - 4\  by\  -2p ^2 - 3p +4

We have to reduce the equation.

According to the question,

To reduce the equation means we need to subtract one equation from the other.

To determine the reduced form of the equation following all the steps given below.

Equation; \rm  2p^ 2 + 3p - 4\  by\  -2p ^2 - 3p +4

Subtraction equation 1 from equation 2,

\rm  = \rm  2p^ 2 + 3p - 4\  -(-2p ^2 - 3p + 4)\\\\=  2p^ 2 + 3p - 4\  -(-2p ^2 -3p +4)\\\\ = 2p^ 2 + 3p - 4+2p^ 2 + 3p - 4 \\\\ = 4p^ 2 + 6p - 8

Hence, The equation in the reduced form of the equation is \rm 4p^ 2 + 6p - 8.

For more details refer to the link given below.

brainly.com/question/1807316

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Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

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

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

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

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

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.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

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