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2 years ago

No links please ! SA = 2πη2 + 2πPh

Mathematics

1 answer:

Georgia [21]2 years ago

5 0

Answer:

216

Step-by-step explanation:

I believe that is the answer

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Area and explanation thanks

adoni [48]

Area of the rectangle is 18•3= 54
Area of the triangle is (10-3)•(18-6)/2= 7•12/2= 42
Area of the whole figure is 54+42=96

3 0

3 years ago

Read 2 more answers

A large box is packed with 6 rows of 8 boxes of dried cranberries in the bottom layer. The box is about 4 layers high. Estimate

Salsk061 [2.6K]

Answer:

Step-by-step explanation:

The estemit will be 0

5 0

2 years ago

Divide 1/7 divided by 6

ArbitrLikvidat [17]

Answer:

1/42

Step-by-step explanation:

1/7 ÷ 6/1

1 x 1 = 1

------------

7 x 6 = 42

1/42

already simplified to the fullest

hope that helps!

6 0

2 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;

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