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4 years ago

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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A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into t

inn [45]

Answer:

The number A(t) of pounds of salt in the tank at time 't' is;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$

Step-by-step explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank, = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

$The \ rate \ of \ change \ in \ A(t) \ with \ time = \dfrac{dA}{dt}$

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

$\dfrac{dA}{dt} = 21 - \dfrac{A(t)}{100}$

Therefore;

$\dfrac{dA}{dt} + \dfrac{A(t)}{100}= 21$

The integrating factor is $e^{\int\limits {\frac{1}{100} } \, dx } = e^{\frac{x}{100} }$

$e^{\dfrac{t}{100} } \times \dfrac{dA}{dt} + e^{\dfrac{t}{100} } \times\dfrac{A(t)}{100}= e^{\dfrac{t}{100} } \times21$

$\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21$

Using an online tool, we get;

${A(t)}= c_1 \times e^{\dfrac{-t}{100} } + 2,100$

At time t = 0, A(t) = 0

We get;

$0= c_1 \times e^{\dfrac{-0}{100} } + 2,100$

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$

8 0

3 years ago

Is (-2, 0) the solution to the system below?

o-na [289]

Answer:

No, the answer is (-3/7, -33/7)

Step-by-step explanation:

y=-3x-6 can be rewritten as y+3x=-6

so, our two equations are:

1. y+3x=-6

2. 3y+2x=-15

we can multiply the first equation by 3.

1-1.3y+9x=-18

Now we can subtract the second equation from equation 1-1.

3y+9x=-18

-(3y+2x=-15)

7x=-3

x=-3/7

Now that we have x, let's plug it into an equation. I will plug it into the first equation.

y=-3(-3/7) - 6

y=9/7 - 6

y=-33/7

so the points are (-3/7, -33/7)

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3 years ago

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snow_tiger [21]

Answer:

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3 years ago

The quadratic equation y = –6x2 + 100x – 180 models the store’s daily profit, y, for selling soccer balls at x dollars. The quad

Mrac [35]

Sample response:

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navik [9.2K]

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