Two-thirds of the money in my pocket is 50 cents.

Mathematics

1 answer:

ch4aika [34]3 years ago

6 0

a.) Two-thirds

b.) 50 cents.

(◍•ᴗ•◍)

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There are three balls for 40 cents or six balls for 60 cents. What percent dose he save by buying one set of six ball instead of

dsp73

Answer:

The persons saves 25% by buying one set of six ball instead of two sets of three balls.

Step-by-step explanation:

Cost of 1 set of three balls = 40 cents

Cost of 2 sets of three balls = 40 cents × 2 = 80 cents

Cost of 1 set of six balls = 60 cents

Percent dose he save by buying one set of six ball instead of two sets of three balls :

= $\frac{80 cents-60 cents}{80 cents}\times 100=25\%$

The persons saves 25% by buying one set of six ball instead of two sets of three balls.

3 0

3 years ago

Solve differential equation:<br><br> 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

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

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