Suppose the length of the rectangle is equal to twice the width, or 2w. If the width is 3, what is the perimeter?

If a rectangle has its length8cm and breadth6cm, then what will be the length of the diagonal of the rectangle?

MaRussiya [10]

Answer:

100

Step-by-step explanation

the width squared plus the height squared

which is 8*8+6*6

=64+36

=100

7 0

3 years ago

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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 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

If a standard drinking straw is 19.5 cm long and has a diameter of 0.6 cm how many square centimeters of plastic are used in one

Liula [17]

The given values of this problem are:
Length = 19. 5 cm
Width (diameter) = 0.6 cm

The unknown is: how much square cm of plastic are used in one straw? In other words the area of the straw.

Derivations and solutions:

Area = length x width
Area = 19. 5 cm x 0.6 cm

Area = 11.7 cm(squared) / 11.7 cm^2

8 0

3 years ago

What does s equal?pls help me plsss.

Archy [21]

Answer:

s< -7/11

Step-by-step explanation:

6 0

3 years ago

6.62 X 0.7= 2.68 X 1.35

Kobotan [32]

Answer:

3.934 = 3.634; hope this helps........

Step-by-step explanation:

Solve:

Multiply the numbers

5.62 ⋅ 0.7 = 2.68 ⋅ 1.35

3.934 = 2.68 ⋅ 1.35

Multiply the numbers

3.934 = 2.68 ⋅ 1.35

3.934 = 3.618

Result

3.934 = 3.618

The input is a contradiction: it has no solutions

5 0

4 years ago

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