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crimeas [40]
3 years ago
11

This is for middle school 8th grade new question need help last one :)

Mathematics
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

He Iransition 1 2 34 y= $ 50 -3.75 (x) y= $30 - 19.99 (x) 7

Step-by-step explanation:

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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
3 years ago
Inequality Match Up Lab
Artist 52 [7]
1. 2x+3<-3 = x<-3
You get this by subtracting 3 from both sides, which gives you -6. Then divide 2 from both sides.

2. -3x<-6 = x>2
You divide -3 from both sides. You must switch the sign because any time you multiply or divide a negative integer from both sides of an inequality the sign is switched.

3. -5x<5 = x>1
See number 2

4. 5-2x<11 = x>-3
Subtract 5 from both sides to get -2x<6. Then, divide both sides by -2 and switch the sign.

5. x+2<5 = x<3
Subtract 2 from both sides

6. 2-3x<5 = x>-1
Subtract 2 from both sides to get -3x<3. Then, divide -3 from both sides and switch the sign.
5 0
2 years ago
Read 2 more answers
What is the measure of CED?<br> O<br> 106°<br> B. 160°<br> O 108°<br> 52<br> 148<br> O 212
LenKa [72]

Answer:

The answer is D (212 degrees)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Factor the expression 100n squared - 1
frez [133]
Factor using the difference of squares.

a^2-b^2 = (a+b)(a-b)

100n^2-1
(10n+1)(10n-1)

Final answer: (10n+1)(10n-1)
4 0
2 years ago
-2-(-4)-2 is equal to what
Juli2301 [7.4K]

the answer is 0 i used a calculator

7 0
1 year ago
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