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3 years ago

Circumference of a circle

Mathematics

1 answer:

solmaris [256]3 years ago

3 0

Answer:

50.27 units

Step-by-step explanation:

The formula for the circumference of a circle is:

C = $\pi$ d

where, d = diameter

Now plug in the values as shown below:

C = $\pi$ 16

C = 50.27

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In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three

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Answer:

The greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

Step-by-step explanation:

We can think of each displacement as a vector, with a given magnitude and direction.

A vector can be written using its x and y coordinates like this

$\vec{t}=(x, y)$

So, for the displacements a and c their vector coordinates would be:

$\vec{a}=(2, 0)$

$\vec{c}=(-1, 0)$

As the b displacement has an angle of 30° toward the north from due east, we can find its x and y coordinates using the following formulas:

$x=(magnitude)*cos(angle)$

$y=(magnitude)*sin(angle)$

Note: the angle in the formula is the one formed with the east measured counterclockwise.

So, the x and y coordinates for the b displacement will be:

$\vec{b}=(2*cos(30), 2*sin(30))$

As the vector additon is commutative, the order won't affect the final position. Nevertheless, any change in the direction of any displacement will change the final position. So, in order to find the combination greatest distance we should calculate the following additions and find the one with the greatest magnitude:

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

Each resultant vector can be found adding each component. Afterwards, the magnitude can be found using the following formula:

$|\vec{R}|=\sqrt[ ]{(R_{x})^2 +{(R_{y})^2}}$

Now, let's calculate!

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$R_{1_x}} =2+2*cos(30)-1=2.73$

$R_{1_y}} =0+2*sin(30)+0=1$

$\vec{R_{1}}=(2.73,1)$

$|\vec{R_{1}}|=\sqrt[ ]{(2.73)^2 +{(1)^2}}=3.86$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$R_{2_x}} =2-2*cos(30)-1=0.73$

$R_{2_y}} =0-2*sin(30)+0=-1$

$\vec{R_{2}}=(-0.73,-1)$ }

$|\vec{R_{2}}|=\sqrt[ ]{(-0.73)^2 +{(-1)^2}}=1.03$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$R_{3_x}} =2+2*cos(30)+1=4.73$

$R_{3_y}} =0+2*sin(30)-0=1$

$\vec{R_{3}}=(4.73,1)$

$|\vec{R_{3}}|=\sqrt[ ]{(4.73)^2 +{(1)^2}}=6.69$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

$R_{4_x}} =2-2*cos(30)+1=1.26$

$R_{4_y}} =0-2*sin(30)-0=1$

$\vec{R_{4}}=(1.26,-1)$

$|\vec{R_{4}}|=\sqrt[ ]{(1.26)^2 +{(-1)^2}}=1.79$

So, after all the calculation, we can know for sure that the vector $\vec{R_{3}}$ has the biggest magnitude. Then, the greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

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3 years ago

Aarons wants to plant an herb garden that is

olya-2409 [2.1K]

Answer:

3 1/3 yards wide

Step-by-step explanation:

A=LW

5= 1.5W

5/1.5= 3.33333 or 3 1/3

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3 years ago