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1 year ago

Which is the value of this expression when m-3 and n=-5

Mathematics

1 answer:

Vedmedyk [2.9K]1 year ago

3 0

Answer:

1/8

Step-by-step explanation:

(6m^-1n^0)^-3 =

= (6 × 1/3 × (-5)^0)^-3

= (6/3)^-3

= 2^-3

= 1/2^3

= 1/8

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What is the volume of a sphere with a radius of 12 inches? Use 3.14 for x.

likoan [24]

the volume of a sphere with a radius of 12 inches is 602.88

8 0

3 years ago

[10 points] A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimens

MissTica

Answer:

6 inches square by 3 inches high

Step-by-step explanation:

For a given surface area, the volume of an open-top box is maximized when it has the shape of half a cube. If the area were than of the whole cube, it would be 216 in² = 6×36 in².

That is, the bottom is 6 inches square, and the sides are 3 inches high.

_____

Let x and h represent the base edge length and box height, respectively. Then we have ...

x² +4xh = 108 . . . . box surface area

Solving for height, we get ...

h = (108 -x²)/(4x) = 27/x -x/4

The volume is the product of base area and height, so is ...

V = x²h = x²(27/x -x/4) = 27x -x³/4

We want to maximize the volume, so we want to set its derivative to zero.

dV/dx = 0 = 27 -(3/4)x²

x² = (4/3)(27) = 36

x = 6

h = 108/x² = 3

The box is 6 inches square and 3 inches high.

_____

Comment on maximum volume, minimum area

In the general case of an open-top box, the volume is maximized when the cost of the bottom and the cost of each pair of opposite sides is the same. Here, the "cost" is simply the area, so the area of the bottom is 1/3 the total area, 36 in².

If the box has a closed top, then each pair of opposite sides will have the same cost for a maximum-volume box. If costs are uniform, the box is a cube.

7 0

3 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

To find the interval on which the series converges you need to use the Ratio Test that says

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago

Aaron bought a set of plastic ice cubes with a mixture of water and air inside.

adoni [48]

Answer:

The volume of water and air inside the plastic ice cube is 3000 mm³

Step-by-step explanation:

* Lets described the figure

- It consists of two identical rectangular pyramids stuck together

in their bases

- The dimensions of the base are 15 mm and 20 mm

- The height of ice cube is 30 mm

∴ The height of each pyramid = 30 ÷ 2 = 15 mm

* Lets talk about the volume of the rectangular pyramid

- The pyramid has a rectangular base and 4 triangular faces

- We have formulas to calculate the volume of the pyramid.

- To find the volume, we use the formula V = (1/3)AH, where

A = area of the pyramid's base and H = height of the pyramid

∵ The dimensions of the base are 20 mm and 15 mm

∴ A = 20 × 15 = 300 mm²

- The height of the pyramid is 15 mm

∵ H = 15 mm

∵ V = 1/3(AH)

∴ V = 1/3(300 × 15) = 1500 mm³

- The ice cube made from two identical rectangular pyramids

∴ The volume of the ice cube = 2 × 1500 = 3000 mm³

- The volume of the water and the air inside the ice cube equal

the volume of the ice cube

* The volume of water and air inside the plastic ice cube is 3000 mm³

3 0

3 years ago

40% of N is 80. Find N

Aliun [14]

Answer:

i think N=32

Step-by-step explanation:

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3 years ago