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3 years ago

HELP ANYONE BRAINLIEST

Mathematics

2 answers:

k0ka [10]3 years ago

7 0

Answer:

12 cups of mild

Step-by-step explanation:

3x3 is 9 and 3/4x9 is 12/4 so 3 9+3 is 12

levacccp [35]3 years ago

6 0

3 3/4*3=11.25
11.24= 11 1/4

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SO anyways :))))

artcher [175]

Answer:

1 / 8², B

Step-by-step explanation:

8 / 8³

8³ = 8*8*8

So, the equation can also be written as:

8 / 8*8*8

divide the numerator and denominator by 8.

1 / 8*8

This is equal to:

1 / 8²

The answer is B.

7 0

2 years ago

use the guide to construct a two column proof proving △aec≅△deb, given that ca is parallel to db and e is the midpoint of ad. gi

MrMuchimi

By using the AAA congruence property of Triangles, △aec≅△deb is proved

It is given two triangles, Δaec and Δdeb, where e is the common point known as the midpoint of ad

Also, it is given that,

ca ║db

We need to prove that, △aec≅△deb

Then we'll use AAA congruence property of Triangles to prove the situation

As ca ║db

then, ∠cae = ∠ebd (Alternate angles)

∠ace = ∠edb (Alternate angles)

and ∠aec = ∠deb (common angles)

Thus, by AAA congruence property, △aec≅△deb

Hence, proved

To learn more about, congruence property, here

brainly.com/question/2039214

#SPJ4

5 0

1 year 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

Como medir la altura de un arbol usando el teorema de thales

zaharov [31]

Answer:

????

Step-by-step explanation:

5 0

3 years ago

192+4(7a+6) Pls Help Show ur work (No spam)

horrorfan [7]

Answer:

192 + 4( 7a + 6 )

DISTRIBUTE

192 + 28a + 24

ADD THE NUMBERS

216 + 28a or 28a + 216

Step-by-step explanation:

U DON'T HAVE TO TYPE "DISTRIBUTE" and "ADD THE NUMBERS"

7 0

2 years ago