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2 years ago

14. 7 + y ( 2y + 6) - 2y?

Mathematics

2 answers:

marusya05 [52]2 years ago

4 0

Answer:

2y²-2y.?+6y+98 esas es la respuesta . espero que te halla servido.

Leviafan [203]2 years ago

4 0

Answer:

2y ^{2} + 4y + 7

Step-by-step explanation:

$7 + y(2y + 6) - 2y ^{2} \\ 7 + 2y ^{2} + 6y - 2y \\ 7 + 2y ^{2} + 4y \\ 2y ^{2} + 4y + 7$

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Whats the gcf of n^3t^2 and nt^4

melomori [17]

$GCF-Greatest\ Common\ Factor\\\\n^3t^2=nt^2\cdot n^2\\nt^4=nt^2\cdot t^2\\\\GCF(n^3t^2;\ nt^4)=\boxed{nt^2}$

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2 years ago

N = 2, a = 16 "Find the indicated real nth root(s) of a.

o-na [289]

X^n = a
x^2 = 16
Square root of 16 = 4
The root is 4

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2 years ago

Solve differential equation: y'''+4y''-16y'-64y=0 y(0)=0, y'(0)=26, y''(0)=-16

Ipatiy [6.2K]

Answer: The required solution of the given differential equation is

$y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.$

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

$y^{\prime\prime\prime}+4y^{\prime\prime}-16y^\prime-64y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\\y(0)=0,~y^\prime(0)=26,~y^{\prim\prime}(0)=-16.$

Let, $y=e^{mx}$ be an auxiliary solution of equation (i).

Then, $y^\prime=me^{mx},~~y^{\prime\prime}=m^2e^{mx},~~y^{\prime\prime\prime}=m^3e^{mx}.$

Substituting these values in equation (i), we get

$m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.$

So, the general solution is given by

$y(x)=Ae^{4x}+Be^{-4x}+Cxe^{-4x}.$

Then, we have

$y^\prime=4Ae^{4x}-4Be^{-4x}-4Cxe^{-4x}+Ce^{-4x},\\\\y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-4Ce^{-4x}-4Ce^{-4x}\\\\\Rightarrow y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-8Ce^{-4x}.$

With the conditions given, we get

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

$y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

and

$y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.$

From equation (iii), we get

$C=26-8A\\\\\Rightarrow 2=26-8A\\\\\Rightarrow 8A=24\\\\\Rightarrow A=3.$

From equation (ii), we get

$B=-3.$

Therefore, the required solution of the given differential equation is

$y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.$

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3 years ago

During swim practice, an Olympic swimmer is required to complete three swimming intervals, in this order: warm-up freestyle, spr

Reika [66]

Answer:

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2 years ago

How do i find 3/4 of 460

liberstina [14]

To find 3/4 of 460 divide 460 by 3/4

5 0

3 years ago

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