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3 years ago

30 POINTS

Mathematics

2 answers:

PtichkaEL [24]3 years ago

6 0

Answer:

180 degrees per second.

Step-by-step explanation:

We have been given a graph that shows the distance covered by a tire or angular velocity of a tire with respect to the number of seconds (x).

Since we know that a tire is a circle and the measure of all the angles of a circle equals to 360 degrees.

We can see from our given graph that the tire completes one rotation in 2 seconds (2.5-0.5), so the distance covered by tire in two seconds will be 360 degrees. This means that the tire turns 360 degrees in each 2 seconds.

$\text{Degrees turned by tire in 2 seconds}=\frac{360^o}{2\text{ seconds}}$

As we are asked to find the degree change by tire per second so we will find the unit rate such that we will have 1 in our denominator. By dividing our numerator and denominator by 2 we will get,

$\text{Degrees turned by tire in every second}=\frac{180^o}{\text{ second}}$

Therefore, the tire is turning 180 degrees per second.

Klio2033 [76]3 years ago

5 0

It takes 2 seconds for the tire to make a full revolution. This is evident from the plot - the curve has a period of length 2.

A full revolution of the tire is a turn of 360 degrees.

So the tire turns at a rate of (360 degrees)/(2 seconds) = 180 degrees/second.

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Express 120 as a product of its prime factors.

raketka [301]

By the Fundamental Theorem of Arithmetic, all number can be expressed as a product of prime numbers.

So naturally, lets divide 120 by an easy prime number.
We know that 120 is even, so lets try 2 120/2 = 60 lets keep dividing it by two until it becomes odd or prime 60/2 = 30 30/2 = 15 now lets see, what are some factors of 15?

Well the obvious ones are 3 and 5, both of which are prime. So now we can just count up how many times we divided it by 2

120/2 = 60 60/2 = 30 30/2 = 15 and 15 is just 3 x 5, so: 
120=(23)×(3)×(5)
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3 years ago

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Lilit [14]

Yes, because there are common factors in the equation. therefore showing you can group

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Is the circumference of all circles about 3 times greater than their diameters. and why?

stira [4]

Answer:

<h2>Yes the circumference is about times greater than their diameters</h2>

Step-by-step explanation:

We will approach this problem by doing a check

Say the diameter of a circle is 4 cm

Hence the radius is 3 cm

We know that the circumference is $C=2\pi r$

$C=2*3.142*2= 12.57cm$

The circumference is about 3 time bigger than the diameter

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3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

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Vikki [24]

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