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trasher [3.6K]
4 years ago
13

Need help don’t know what to do I’m confused

Mathematics
2 answers:
Artemon [7]4 years ago
7 0
The last one is the answer h=2t-d distribute the 1/2 into the (d + h) you get t=1/2d + 1/2 h then multiply everything by the denominator of 2 this would give you 2t = d + h because the denominator of 2 on the d and h cancel out and leaves a 1 infront which is just d and h then i’d u switched the h and 2t across the equation they become the opposite sign -h = -2t + d since you don’t want -h switch all the signs leaving you with the answer h = 2t - d
ankoles [38]4 years ago
4 0

Answer:

last option;  h = 2t - d

Step-by-step explanation:

t = 1/2(d + h)

t = \frac{d}{2} + \frac{h}{2}

now multiply each side by 2 to eliminate denominator

2t = 2 (\frac{d}{2} + \frac{h}{2})

2t = d + h

h = 2t - d

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5 0
4 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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