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2 years ago

The slope of the lines that passes through the points (-10,0) and (-13,3)

Mathematics

1 answer:

goldenfox [79]2 years ago

4 0

Answer:

-1

Step-by-step explanation:

The formula to find the slope of a line is:

$\frac{y_{2}- y_{1} }{x_{2} -x_{1} }$

where (x1, y1) and (x2, y2) are points on the line. We can substitute the points (-10, 0) and (-13, 3) into the formula and simplify:

$\frac{3-0}{-13-(-10)} =\frac{3}{-13+10} =\frac{3}{-3} =-1$

This means the slope of the line is -1.

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lisabon 2012 [21]

Answer:

He multiplied 25 by 1 and then multiplied 25 by 9 correctly. He added the two partial products correctly

Step-by-step explanation:

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2 years ago

Elizabeth has 1 7/10 as many plants as Rosalie has in her arden. If Elizabeth has 51 plants, how many plants does Rosalie have i

dexar [7]

Question: Elizabeth has 1 7/10 as many plants as Rosalie has in her garden. If Elizabeth has 51 plants, how many plants does Rosalie have in her garden?

Answer: X=30

explanation:

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question answered by

(jacemorris04)

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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:

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

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3 years ago

Determine how many lines of symmetry each object has. Then determine weather each object has rotational symmetry

Dafna1 [17]

Answer:

2: yes

Step-by-step explanation:

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3 years ago

How do u find the perimeter of somthing

Kazeer [188]

Lets say that the length of a rectangle is 12 cm and the width is 5 cm, you would add 12+12+5+5 (as in the length on both length sides and the width on both sides) so the perimeter of this rectangle is: 34 cm

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2 years ago

Read 2 more answers