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3 years ago

8

The distance from the center of a computer disc to the edge of the disc is 6 centimeters. What is the area of the compact disc?

A. 113.04 B. 18.84 C. 36 D. 452.16

2 answers:

valina [46]3 years ago

8 0

C es equal to 36 because that shape make a square

mojhsa [17]3 years ago

5 0

The answer to your question is #36

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nikklg [1K]

The y intercept is where the line hits the verticals line of the graph. Hope this helps!

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2 years ago

A circle has a radius of 6cm . find the length s of the arc intercepted by a central angle of 169° .

Igoryamba

The circumference of the circle is = 2 x pi x 6 = 37.7cm

To find the arc = 360 divided by 169 = 2.13

The arc = 37.7 divided by 2.13 = 17.69 cm

3 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 <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> 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 <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 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) <em>C</em> 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 <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 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

2 years ago

8% of the employees at a shop work part-time.

kifflom [539]

Answer:50

Step-by-step explanation:

7 0

2 years ago

The function g(x) = 8x2 – 48x + 65 written in vertex form is g(x) = 8(x – 3)2 – 7. What is the vertex of g(x)?

maksim [4K]

The answer is (3, -7). If the function is written in the form y = a(x – h)^2 + k, the vertex will be (h, k). Let's write the function 8x^2 – 48x + 65 in the form of a(x – h)^2 + k. g(x) = 8x^2 – 48x + 65. g(x) = 8x^2 – 48x + 72 - 72 + 65. g(x) = (8x^2 – 48x + 72) - 7. g(x) = (8 * x^2 – 8 * 6x + 8 * 9) - 7. g(x) = 8(x^2 - 6x + 9) - 7. g(x) = 8(x - 3)^2 - 7. The function is now in the form a(x – h)^2 + k, where a = 8, h = 3, and k = -7. Thus, the vertex is (3, -7).

4 0

3 years ago

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