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3 years ago

Help me out,With that question

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

B(boxes) < 3 1/2 so this means that any box less than 3 1/2 can fit through the doorway

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Find the circumference and area of each circle 15mn

AysviL [449]

Area= 709.86
Circumference= 94.25

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3 years ago

The graph shows the rate at which paint is used to paint a wall. wat are the values to each statement below based on the graph.

zhuklara [117]

Answer:

Step-by-step explanation:

When 1 gallon is has been used 400 square feet of the wall will be paint
The unit rate is 400 square feet per gallon

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3 years ago

A $20 shirt was 30% off and you had a coupon for another 10% off. What is the new price?

laila [671]

Answer:

$12.60 is the new price!

Step-by-step explanation:

20 x 0.7 = 14

14 x 0.9 = 12.6

Hope this helped!

4 0

3 years ago

Read 2 more answers

use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle

Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

$\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

$\int_C9y^3dx-9x^3dy$

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

$\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2$

$\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2$

$\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy$

⇒ $-27\int\int_D(x^2 + y^2)dx dy$

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

$-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta$

⇒ $-27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta$

⇒ $-27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta$

⇒ $-27\int\limits^{2\pi}_0 (16/4) d\theta$

⇒ $-108\int\limits^{2\pi}_0 d\theta$

⇒ $-108[2\pi - 0]$

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0

2 years ago

Need Help With Math Problem

ioda

It’s b i’ve had this before

7 0

2 years ago