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2 years ago

Math Story problem. I need help

Mathematics

1 answer:

Olin [163]2 years ago

6 0

Answer:

i believe it would be 12

hope this helps !!

Step-by-step explanation:

96 divided by 8 would equal 12 so therefore the answer is 12

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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

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

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

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

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

4 0

3 years ago

CDEF is a kite and m

yan [13]

What? i don’t understand

3 0

3 years ago

Equation of a circle whose center is at (4,0) and radius is length 2/3

Masteriza [31]

We need (x-h)^2 + (y-k)^2 = r^2

h = 4, k = 0 and r = 2/3

Take it from here.

5 0

2 years ago

Question 12 (2 points)

elena55 [62]

Answer:

A box plot is drawn with end points at 24 and 49.The box extends from 28 to 44 and a vertical line is drawn inside the box at 34.

Step-by-step explanation:

Ordering the data given :

24,28,32,34,40,44,49

We can calculate the 5 number summary required to give the appropriate boxplot that can be produced :

Minimum = 24

Maximum = 49

Median = 1/2(n+1)th term

n = 7

Median = 1/2(8) = 4th term

Median = 34

Lower quartile, Q1 = 1/4(n+1)th term

n = 7

1/4(8) = 2nd term

Q1 = 28

Upper quartile : 3/4(n+1)th term

n = 7

Q3 = 3/4(8) = 6th term

Q3= 44

3 0

2 years ago

Can you guys help me with this one please

PilotLPTM [1.2K]

Answer:

1 to 100,000

Step-by-step explanation:

1 km is equivalent to 100,000 cm.

3 0

2 years ago