All categories

3 years ago

Please help

Mathematics

2 answers:

blondinia [14]3 years ago

7 0

Answer:

D. 7200

Step-by-step explanation:

Hunter-Best [27]3 years ago

3 0

Statement:

A softball field is a square with sides of length 60 feet.

To find out:

The shortest distance between the first base and third base, i.e., the displacement

Solution:

Let us consider 1st base as A, Home Plate as B and 3rd base as C.
AB = 60 ft
BC = 60 ft.
To find out the shortest distance between the first base and third base we have to find out the value of AC.
In right triangle ABC,
(AB)² + (BC)² = (AC)² [By Pythagoras Theorem]
or, (AC)² = (60 ft)² + (60 ft)²
or, (AC)² = 3600 ft² + 3600 ft²
or, (AC)² = 7200 ft²
or, AC = √7200 ft
So, the shortest distance will be √7200 ft.

Answer:

D. √7200 ft

Hope you could understand.

If you have any query, feel free to ask.

You might be interested in

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;

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