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2 years ago

What is the word for: The solid, liquid, or gas through which sound travels:

Mathematics

2 answers:

DIA [1.3K]2 years ago

7 0

Answer:

A material through which sound waves can travel is called a medium.

mash [69]2 years ago

6 0

Answer:

I believe your answer would be Soundis

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Simplify. (-2xy4)^4

IrinaVladis [17]

Step-by-step explanation:

${( - 2x {y}^{4} })^{4}$

$= {( - 2)}^{4} \times {x}^{4} \times { {y}^{4} }^{4}$

$= 16 {x}^{4} {y}^{16}$

3 0

3 years ago

Pls help ASAP NUMBER 2

Lisa [10]

Answer:

The answer to the question is 2 and 5

5 0

3 years ago

Read 2 more answers

A circle with an area 9pi has a sector with a central angle of 1/9pi what is the area of the sector

ludmilkaskok [199]

Answer:

½pi or 0.5pi units²

Step-by-step explanation:

Area : angle

9pi : 2pi

x : pi/9

9pi/2pi = x/(pi/9)

4.5 × pi/9 = x

x = ½pi or 0.5pi units²

8 0

3 years ago

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago

A cruise ship is traveling at a speed of 20 knots . A knot is approximately equal to 1.151 miles per hour. What is the approxima

Mrrafil [7]

1 mile = 1760 yards
20 * 1.151 = 23.02
23.02 * 1760 = 40,515.2
40,515.2 / 60 = 675.243~
675.243~ / 60 = 11.2542~
They are traveling 11.25 yards per second
I took the number of knots multiplied by the approximate miles per hour equal
Then multiplied that by the number of yards in a mile, and then divided that by the number of minutes in an hour, and divided that by the number of seconds in a minute. That is how I got the answer, I hope it helps!

6 0

3 years ago