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3 years ago

Which point is graphed on the coordinate plane?

Mathematics

2 answers:

Eduardwww [97]3 years ago

5 0

Answer:

8,3

Step-by-step explanation:

xz_007 [3.2K]3 years ago

3 0

Answer:

(8, 3)

Step-by-step explanation:

You can see that it is moved 8 places to the right which represents x (the first number in your coordinate pair). It also moves 3 places above which represents the y (the second number in the coordinate pair).

You can always remember that x goes before y when writing a coordinate pair by the alphabetical order x, y, z..- X goes before Y!

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10p-2(3p-6)=4(5p-6)-10p

zvonat [6]

$10p-2(3p-6)=4(5p-6)-10p\ \ \ \ | omiting\ brackets\\\\ 10p-6p+12=20p-24-10p\\\\
4p+12=10p-24\ \ \ \ | subtract\ 10p\\\\
-6p+12=-24\ \ \ | subtract\ 12\\\\
-6p=-36\ \ \ \ | divide\ by\ -6\\\\
p=6\\\\Solution\ is\ p=6.$

5 0

3 years ago

Can someone please give me the answers (I’ll be giving brainliest)

Andre45 [30]

I think it's the first three.
-7.5 < -7
-4 < -3
-0.5 < -1
(those are the inequalities after i put each of the values for x in the original equation)

8 0

2 years ago

5 plus 2 minus 7 plus 4

Yakvenalex [24]

Answer:

5+2=7-7+4=4 =4 the answer is 4

Step-by-step explanation:

do not cheat 1st grader

5 0

3 years ago

Read 2 more answers

ABCD is a rhombus in which angle BAD = 68°

olchik [2.2K]

Answer:

21 69 34

Step-by-step explanation:

you welcome

8 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