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3 years ago

9

Parse the regular hexagon, and tell how much regular unit triangle do you find, if the length of one side of the regular hexagon

is n unit(s):

1 answer:

Thepotemich [5.8K]3 years ago

5 0

Answer:

Six.

Step-by-step explanation:

In geometry, a hexagon is a two-dimensional polygon that has six sides. A regular hexagon is a hexagon in which all of its sides have equal length. We sometimes define a regular hexagon using equilateral triangles, or triangles in which all of the sides have equal length.

The regular hexagon is a convex polygon with six equal sides and six equal angles. Each external angle of the regular hexagon measures 60 degrees. It is closely related to equilateral triangles: Joining each vertex with its opposite, the regular hexagon is divided into six equilateral triangles.

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If a line falls on the points (6,22) and (4,21), what is its slope? Enter your answer as a fraction in lowest terms. Use a slash

I am Lyosha [343]

Answer:

1/2

Step-by-step explanation:

Since we have two points, we can use the slope formula

m = (y2-y1)/(x2-x1)

= (22-21)/(6-4)

= 1/2

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The answer is principal.
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Could somebody please answer these four questions with simple yet explanatory answers?

Kamila [148]

1) Surface Area = 2(lw + lh + wh)
2(4*5+4*9+5*9) =202in^2
Surface Area=202in^2
Lateral Area= Perimeter of base * height
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4+4=8 <---length + length
10+8=18 <--total perimeter
18 * 9=162in^2 <--multiplied the height +perimeter of base
Lateral Area=162in^2

2) Same concept as the previous one
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2(4*2+4*5+5*2) =76in^2
Surface Area= 76in^2

Lateral Area
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Morgarella [4.7K]

Answer:

C) (x-3)(x+3)

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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 :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

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

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

which is a family of ellipses.

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

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

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 $\frac{dr}{dt}=\lambda \triangledown C$

$\Rightarrow \frac{dx}{dt}=-\frac{2\lambda x}{10^4}e^{-(x^2+2y^2)/10^4}, \frac{dy}{dt}=-\frac{4\lambda y}{10^4}e^{-(x^2+2y^2)/10^4}$

∴ $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2y}{x}$

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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3 years ago