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3 years ago

Find the area of a semicircle

Mathematics

2 answers:

Furkat [3]3 years ago

6 0

Answer:

18pi, or 56.52

Step-by-step explanation:

To start off, if you were to find the area of any whole circle, you would use the formula: pi x radius^2 = area of circle

However, since this is a semi-circle (half of a whole circle) you would divide the whole expression by 2, and then solve:

(pi x radius^2)/2 = area of semi-circle

(pi x 6^2)/2 = area of semi-circle

(36 x pi)/2 = area of semi-circle

18 x pi = area of semi-circle

18 x 3.14 = area of semi-circle

56.52 = area of semi-circle

Inessa [10]3 years ago

6 0

The area of a circle is

$\pi \: {r}^{2}$

So the area of a semicircle would be half of that, or

$\frac{\pi}{2} {r}^{2}$

So just plug in the radius, six, to get this

$18\pi$

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A manufacturer of shipping boxes has a box shaped like a cube. The side length is

Debora [2.8K]

The vol. of a cube of side length s is V=s^3.

So, if the side length is 5a+4b, then the volume of the cube is

V(a, b) = (5a+4b)^3. This could, of course, be expanded, using the binomial theorem, but there's no point in doing so.

7 0

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

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

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

7 0

3 years ago

The graph shows the distance a bicyclist traveled from a starting point in 10-second increments. Which of these is closest to th

myrzilka [38]

Answer:

C. 10 m/s

Step-by-step explanation:

3 0

3 years ago

Wendy has 5 and 1 over 3 feet of ribbon. How many 2 over 3 foot pieces can Wendy cut from the 5 and 1 over 3 feet of ribbon? Sho

denis-greek [22]

Answer:

Your answer is 8

Step-by-step explanation:

First you have to convert the mixed number into an improper fraction. To do this, multiply the denominator by the whole number. Then add the numerator and put that answer in a fraction over the original. So it would be 3*5=15+1=16/3. So now that the denominators are the same, all you need to do is see how many times 2 goes into 16. 16/2 which is 8. Answer = 8

6 0

3 years ago

In circle o, the length of radius OL is 6 cm and the length

AlekseyPX

Answer:

14.2cm

Step-by-step explanation:

The diagram representing the circle and its attributes has been attached to this response.

As shown in the diagram;

The circle is centered at o,

The length of radius OL = 6cm

The length of the arc LM = 6.3cm

The angle MON = 75°

The angle LOM = θ

Remember that;

The length, L, of an arc is given by;

L = (θ / 360) x (2πr) -------------(i)

Where;

θ is the angle subtended by the arc

r = radius of the circle.

Using the formula in equation (i), let's calculate the angle θ subtended by arc LM as follows;

L = (θ / 360) x (2πr)

Where;

L = length of arc LM = 6.3cm

r = radius of the circle = length of radius OL = 6cm

Substitute these values into the equation to get;

6.3 = (θ / 360) x (2 x π x 6)

6.3 = (θ / 360) x (12 x π)

6.3 = (θ / 30) x (π) [Take π = 22/7]

6.3 = (θ / 30) x (22 / 7)

θ = $\frac{6.3*30*7}{22}$

θ = 60.14°

Therefore, the angle subtended by arc LM is 60.14°

Now, from the diagram,

The angle subtended by arc LMN is;

θ + 75° = 60.14° + 75° = 135.14°

Let's now calculate the length of arc LMN using the same equation (i)

L = (θ / 360) x (2πr)

Where;

L = length of arc LMN

θ = angle subtended by LMN = 135.14°

r = radius of the circle = length of radius OL = 6cm

Substitute these values into the equation;

L = (135.14° / 360°) x (2 x π x 6) [Take π = 22/7]

L = 14.15cm

Therefore, the length of arc LMN is 14.2cm to the nearest tenth.

6 0

3 years ago