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3 years ago

Whats the name of the geometric figure

Mathematics

1 answer:

sdas [7]3 years ago

3 0

which figure? plz tell us more

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The set of natural numbers between 16 and 156

SIZIF [17.4K]

Answer:

All numbers not decimals are natural numbers

Step-by-step explanation:

4 0

4 years ago

Find X

Dima020 [189]

Answer:

x=10

Step-by-step explanation:

3x+8x=100+10

11x=110

x=110/11

x=10

5 0

3 years ago

Find the area of the shaded region. Round your answer to the nearest tenth.

Troyanec [42]

Area of the square = (9 m)^2 = 81 m^2

The four circular sectors combine to form a circle with diameter 9 m, hence radius 4.5 m.

Area of the circle = π (4.5 m)^2 = 20.25π m^2

Then the area of the shaded region is

81 m^2 - 20.25π m^2 ≈ 17.4 m^2

7 0

3 years ago

what is perpendicular to y= 3x -9 and passes through the point (3,1)

GREYUIT [131]

$(3,1) , \ \ y= 3x -9 \\ \\ The \ slope \ is :m _{1} =3 \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _{1} \cdot m _{2} = -1 \\ \\3\cdot m_{2}=-1 \ \ /:3\\ \\ m_{2}=-\frac{1}{3}$

$\Now \ your \ equation \ of \ line \ passing \ through \ (3,1) would \ be: \\ \\ y=m_{2}x+b \\ \\1=-\frac{1}{ 3} \cdot3 + b$

$1=-1+b\\ \\ b=1+1\\ \\b=2 \\ \\ y =- \frac{1}{3}x+2$

6 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