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mezya [45]
3 years ago
10

Pls help to get brainlist

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
7 0

Answer: 15/10-12/10= 3/10

Step-by-step explanation:

Lubov Fominskaja [6]3 years ago
4 0

Answer: Please give brainliest!!!

\frac{3}{2} = \frac{15}{10}

\frac{15}{10} - \frac{12}{10} = \frac{3}{10}

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Which of the following best describes the slope of the line below? -5 O A. Undefined O B. Zero O c. Positive OD. Negative​
lana66690 [7]
negative is the answer
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Lowest common multiple of 10 and 23
PIT_PIT [208]

Answer:230

Step-by-step explanation:

only one where 23 has multiple ens with 0

4 0
4 years ago
20 + 30 × 0 + 1 =<br>ok... can you answer it now?
vagabundo [1.1K]
<span>20 + 30 × 0 + 1 
= </span><span>20 + 0 + 1 
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6 0
3 years ago
-20 represents the location of a fish 20 feet below sea
Svetllana [295]

Answer:

Sea level

Step-by-step explanation:

That’s what I learned couple weeks ago

8 0
3 years ago
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
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