The first three terms of the binomial expansion of (2 – ax)" are 64 – 16bx +100bx?.
Find the value of each of the integers n, a and b.

Ans: n=6, a=5, b=60

The First Three Terms Of The Binomial Expansion Of (2 Ax)" Are 64 16bx +100bx?.Find The Value Of Each

Answers

Answer 1

Answer:

n = 6, a = 5, b = 60

Step-by-step explanation:

In a binomial function (a + b)ⁿ expression that represents the terms,

(a + b)ⁿ = [tex]\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k[/tex]

By this formula,

1st term = [tex]\binom{n}{0}a^{n-0}b^0[/tex] = aⁿ

2nd term = [tex]\binom{n}{1}a^{n-1}b^1[/tex] = [tex]n.a^{n-1}b^1[/tex]

3rd term = [tex]\binom{n}{2}a^{n-2}b^2[/tex] = [tex]\frac{n(n-1)}{2}.a^{n-2}.b^2[/tex]

For the binomial expansion initial 3 terms of (2 - ax)ⁿ = 64 - 16bx + 100bx²

Terms of (2 - ax)ⁿ = [tex]2^n+n(2)^{n-1}(-ax)+\frac{n(n-1)}{2}(2)^{n-2}(-ax)^2[/tex]

                             = [tex]2^n-n(2)^{n-1}(ax)+n(n-1)(2)^{n-3}(a^2x^2)[/tex]

Comparing the terms of both the expansions,

1st term

2ⁿ = 64

2ⁿ = 2⁶

n = 6

2d term

[tex]n(2)^{n-1}(ax)=16bx[/tex]

[tex]6(2)^{6-1}(a)=16b[/tex]

192a = 16b

b = 12a -----(1)

3rd term

[tex]n(n-1)2^{n-3}(a^2x^2)=100bx^2[/tex]

[tex]6(6-1)2^{(6-3)}(a^2)=100b[/tex]

[tex]30(2)^3(a^2)=100b[/tex]

240a² = 100b

b = 2.4a² -----(2)

From equation (1) and (2),

b = 12a = 2.4a²

a = [tex]\frac{12}{2.4}=5[/tex]

From equation (1)

b = 12a = 60

Therefore, n = 6, a = 5, b = 60


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Answer:

Step-by-step explanation:

1). Swap the order of equations.

2). Yes, system of equations are equivalent with the same solution.

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y = - [tex]\frac{184}{29}[/tex] + [tex]\frac{203}{29}[/tex] = [tex]\frac{19}{29}[/tex]

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Step-by-step explanation:

) In each diagram, line p is parallel to line f, and line t intersects lines p and f. Based
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Answers

Answer:

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Answers

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Answers

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Step-by-step explanation:

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Answers

Answer:

1. 133 1/3% of 36=48

2. 260% of 360=936

Step-by-step explanation:

1. 133 1/3% of 36

=(133 1/3)%  of 36

=(133/1+1/3)% of 36

=(399+1/3)% of 36

=(400/3)% of 36

=(400/3 *100) *36

=(400/300*36)

=4/3*36

=4*12

=48  Ans

---------------------------------------------------------------------------------------

2. 260% of 360

=260% of 360

=260/100*360

=26/10*360

=26/1*36

=936 Ans

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A

Step-by-step explanation:

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Answer:

1st answer choice- You do not have enough jars because 20 jars will only hold 35 cups of applesauce.

Step-by-step explanation:

So 1.75 cups of applesauce fits in a jar. You have 20 jars so you multiply the cups of applesauce that can fit in the jar by how many jars you have. It will look like this: 1.75x20=? Now you multiply them and you get 35 which is how many cups of applesauce 20 jars will hold. You will have 5 cups of applesauce left over that cannot be transformed into a jar.

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Answer:

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Step-by-step explanation:

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i believe its C

Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Which figure can be transformed into figure L by a 90° rotation clockwise about the origin followed by a translation 2 units down?

Question Options:
Question 2 options:

figure J

figure M

figure N

figure P

Answers

Answer:

J can be rotated

Step-by-step explanation:

Answer: Figure J

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Answers

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the slope is 2

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2

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[tex]y = - \frac{1}{6} x - 1 \\ [/tex]

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

A rectangular frame has length (x+2) units and width (x-2) units. If the area is 96 square units, what is the value of x?

Answers

Answer:

x=10

Step-by-step explanation:

according to the question

(x+2)(x-2)=96

=>x^2-4=96

=>x^2=96+4

=>x^2=100

=>x=√100

Therefore

x=10

x=-10

As width and length can't be negative

So x=10

Width=10-2=8

Length=10+2=12

12×8=96 (proven)

Answer:

Please check the explanation.

Step-by-step explanation:

As we know that the area of a rectangle is defined by multiplying the length by the width.

[tex]A=l\times w[/tex]

Given

rectangular frame length = l = (x+2) unitsrectangular frame width = w = (x-2) unitsArea = 96 square units

substituting all the given values in the formula to find the value of x.

[tex]A=l\times w[/tex]

[tex]96=\left(x+2\right)\times \left(x-2\right)[/tex]

[tex]96=x^2-4[/tex]

[tex]x^2-4=96[/tex]

subtract 96 from both sides

[tex]x^2-4-96=96-96[/tex]

[tex]x^2-100=0[/tex]

[tex]x^2=100[/tex]

[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]

[tex]x=\sqrt{100},\:x=-\sqrt{100}[/tex]

[tex]x=10,\:x=-10[/tex]

Putting x = -10 in the length and width will make the length and width negative, which can not be possible.

i.e.

length = l = x+2 = -10+2 = -8 units

width = w = x-2 = -10-2 = -12 units

Therefore, x=-10 must be excluded.

Now, putting the length of x = 10.

i.e.

length = l = 10+2 = 10+2 = 12 units

width = w = x-2 = 10-2 = 8

[tex]A=l\times w[/tex]

96 = 12 × 8

96 = 96

Therefore, the correct value of x = 10

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