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8090 [49]
2 years ago
10

Finding area of shaded sectors.

Mathematics
1 answer:
tekilochka [14]2 years ago
5 0

Answer:

69.81 sq. m. (rounded to 2 decimal places)

Step-by-step explanation:

The sector of a circle is "part" or "portion" of a circle. The formula for the area of a sector is:

A=\frac{\theta}{360}*\pi r^2

Where

\theta  is the central angle

r is the radius

Given the figure, the arc is given as 80 degrees, but not the central angle of the shaded sector. But from geometry we know that the central angle and the intercepted arc have the same measure. So we can say:

\theta = 80

Also, the radius of the circle shown is 10 meters, so

r = 10

Now, we substitute in formula and find our answer:

A=\frac{\theta}{360}*\pi r^2\\A=\frac{80}{360}*\pi (10)^2\\A=\frac{2}{9}*100\pi\\A=\frac{200\pi}{9}\\A=69.81

Thus,

The area of the shaded sector is 69.81 sq. meters.

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2x+5x=100 x intercept​
Papessa [141]

This is an equation involving x alone, so the most you can do is solve it:

2x+5x=100x \iff 7x=100x \iff 0=93x \iff x=0

3 0
3 years ago
The residual value in a car lease agreement is what the leasing company believes the car will be worth at the end of your lease.
jok3333 [9.3K]

Answer:

$10,560

Step-by-step explanation:

48% of $22,000 is $10,560

3 0
3 years ago
Mohamed and Li Jing were asked to find an explicit formula for the sequence -5, -25, -125, -625,....
Nuetrik [128]

Answer:

Li Jing's formula i.e.  \boxed{g_n=-5\cdot \:5^{n-1}}  is right.

Step-by-step explanation:

Considering the sequence

-5,\:-25,\:-125,\:-625,...

A geometric sequence has a constant ratio r and is defined by

g_n=g_0\cdot r^{n-1}

\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{g_{n+1}}{g_n}

\frac{-25}{-5}=5,\:\quad \frac{-125}{-25}=5,\:\quad \frac{-625}{-125}=5

\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}

r=5

So, the sequence is geometric.

as

\mathrm{The\:first\:element\:of\:the\:sequence\:is}

g_1=-5

r=5

so

g_n=g_1\cdot r^{n-1}

\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:

g_n=-5\cdot \:5^{n-1}

Therefore, Li Jing's formula i.e.  \boxed{g_n=-5\cdot \:5^{n-1}}  is right.

8 0
3 years ago
A game decreased in price by 1 5 . After the reduction it was priced at £48. What was the original price of the game?
vovikov84 [41]

Answer:

£63

Step-by-step explanation:

when we subtract 15 from 48 ,we get 63 which is the answer.

8 0
2 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

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

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

7 0
2 years ago
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