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Marizza181 [45]
4 years ago
8

Find the 4th term (x-y)^12

Mathematics
1 answer:
maria [59]4 years ago
3 0

Answer:

The fourth term of the expansion is -220 * x^9 * y^3

Step-by-step explanation:

Question:

Find the fourth term in (x-y)^12

Solution:

Notation: "n choose k", or combination of k objects from n objects,

C(n,k) = n! / ( k! (n-k)! )

For example, C(12,4) = 12! / (4! 8!) = 495

Using the binomial expansion formula

(a+b)^n

= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n

For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is

C(n,3)a^(n-3)b^3

=C(12,3) * x^(12-3) * (-y)^(3)

= 220*x^9*(-y)^3

= -220 * x^9 * y^3

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Answer:

96 feet

Step-by-step explanation:

This problem requires the use of pythagorean theorem.

she walks 24 ft south + 32 ft east + \sqrt{24^2 + 32^2} = 56 + \sqrt{1600} = 56 + 40 = 96

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LMN is an isosceles triangle. What is the approximate length of side LM, and what is the approximate perimeter of triangle LMN?
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Step-by-step explanation:

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Can you help me find the percent?<br><br> From 42 acres to 72 acres
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Percentage change:×100%

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Find a homogeneous linear differential equation with constant coefficients whose general solution is given. y = c1 cos(x) + c2 s
Anton [14]

Answer:

The homogeneous differential equation is

d^4y/dx^4 + 37d²y/dx² + 36 = 0

Or

y^(iv) + 37y'' + 36 = 0

Step-by-step explanation:

We want to find a homogeneous linear differential equation with constant coefficients whose general solution is given as

y = c1 cos(x) + c2 sin(x) + c3 cos(6x) + c4 sin(6x)

This, we are working in reverse. Instead of having a differential equation, and writing and solving it's characteristic equation, before writing it's general solution, we are already given the general solution, so, we work in reverse.

Given

y = c1 cos(x) + c2 sin(x) + c3 cos(6x) + c4 sin(6x)

By observation, we have

(1) Homogeneous differential equation of the fourth order, because of the number of constants

(2) Two pairs of complex roots, that's why we have cosines, and sines.

Knowing that if

m = a ± ib

The general solution is

y = (e^ax)(c1 cos bx + c2 sin bx)

We can now write for

y = [c1 cos(x) + c2 sin(x)] + [c3 cos(6x) + c4 sin(6x)]

As

m = 0 ± i

and

m = 0 ± 6i

So that we have

m = √(-1)

m = √(-36)

m² = -1

m² = -36

m² + 1 = 0

m² + 36 = 0

The auxiliary equation is therefore

(m² + 1)(m² + 36) = 0

m^4 + 36m² + m² + 36 = 0

m^4 + 37m² + 36 = 0

Finally, the homogeneous differential equation is

d^4y/dx^4 + 37d²y/dx² + 36 = 0

Or

y^(iv) + 37y'' + 36 = 0

5 0
3 years ago
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