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mrs_skeptik [129]

2 years ago

8

What is the coefficient of the x9y-term in the binomial expansion of (2y + 4x3)4? 4 32 128 512

2 answers:

sasho [114]2 years ago

7 0

Answer:- 512

Explanation:-

We know that $(m+1)^{th},\ (T_{m+1})$ in the binomial expansion $(p+q)^n$ is given by

$T_{m+1}=^nC_m\ p^{n-m}q^m$

Assume that $x^9y$ occurs in the $(m+1)^{th}$ term of the expansion of $(2y+4x^3)^4=(4x^3+2y)^4$

$T_{m+1}=^4C_m\ (4x^3)^{4-m}(2y)^m$

Comparing power of x and y in $x^9y$ we get m=1

Thus term for m=1 $=\ ^4C_1\ (4x^3)^{3}(2y)^1=^4C_1=\frac{4!}{(4-1)!1!}(64x^9)(2y)=4(128x^9y)=512x^9y$

Thus the coefficient of $x^9y$ is 512.

Nesterboy [21]2 years ago

5 0

512 is the coefficient of the x9y-term

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A company purchased a delivery van for $28,000 with a salvage value of $3,000 on September 1, Year 1. It has an estimated useful

Tems11 [23]

Answer:

Depreciation till December 31, Year 1 will be equal to 1,250$

Step-by-step explanation:

Purchasing Cost = 28,000$

Salvage Value = 3,000$

Total Depreciation:

Total Depreciation over 5 years (60 Months) = Purchasing Cost - Salvage Value

Total Depreciation over 5 years (60 Months) = 28,000 - 3,000

Total Depreciation over 5 years (60 Months) = 25,000$

Monthly Depreciation:

Using the unity method we have monthly depreciation by dividing the total depreciation by the total no. of months as below:

Total Depreciation over a single month =25,000/60

Total Depreciation over a single month = 416.67$ (Monthly Depreciation)

Depreciation till December 31, Year 1

As from September 1, Year 1 to December 31, Year 1, its been 3 months therefore total depreciation will be = 3 * Monthly Depreciation

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2 years ago

terrence made a sphere out of modeling clay. the sphere had a radius of 2 inches. approximately how much modeling clay did terre

navik [9.2K]

Amount of clay used = volume of the sphere

It is given that the radius of the sphere is 2 inches.

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Hence, the amount of modeling clay Terrence used $=\frac{32}{3} \pi$ cubic inches.

5 0

2 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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1 year ago

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Hitman42 [59]

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5 0

1 year ago

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Vedmedyk [2.9K]

Answer:

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Step-by-step explanation:

Volume of any shape is given by cross sectional area multiplied by height. Therefore, V=Ah

Where A is area and h is height

Since their cross sections are the same, then volume depends on height.

If 12cm equals 2 liters

8cm equals x

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Approximayely, the smaller pail has capacity of 1.33 liters

7 0

2 years ago