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3 years ago

7

A scale drawing of a room is shown below. In the drawing, the room is 7 1/2 inches long and 5 inches wide. What is the actual ar

ea of the room? Please help

1 answer:

pashok25 [27]3 years ago

8 0

Answer:

The area of the room is 37.5 square inches.

Step-by-step explanation:

Consider that the room is rectangular in shape.

The dimensions of the rooms are:

Length (l) = 7.5 inches

Breadth (b) = 5 inches

The area of a rectangle is:

Area = l × b

Compute the area of the room as follows:

Area = l × b

$=7.5\times 5\\=37.5$

Thus, the area of the room is 37.5 square inches.

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3 years ago

Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w

siniylev [52]

Solution :

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are actually the curves

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where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

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We sketch the level curves for K =1,2,3 and 4.

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Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.

$r(t)=x(t)i+y(t)j$ is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

So then we have that $\frac{dr}{dt}=\lambda \triangledown C$

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Using separation of variables,

$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

$\Rightarrow kx_0^2=y_0$

$\Rightarrow k =\frac{y_0}{x_0^2}$

∴ The path of the shark will follow is along the parabola

$y=\frac{y_0}{x_0^2}x^2$

$y=y_0\left(\frac{x}{x_0}\right)^2$

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