What is the area of this trapezoid?

KengaRu [80]

Common Factors of 32 and 36
factors of 32 = 1, 2, 4, 8, 16 and 32
factors of 36 = 1, 2, 3, 4, 6, 9, 18, 36
So the common factors of 32 and 36 are 1, 2 and 4.
The greatest common factor is 4

4 0

3 years ago

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Will someone plz give me brainliest :(

nevsk [136]

Answer:

What do you need help with?

6 0

3 years ago

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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

5 0

2 years ago

You roll 2 number cubes. What is the probability of rolling double twos? EZ points!

Misha Larkins [42]

Answer:

Step-by-step explanation:

There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.

3 0

2 years ago

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Jonathan is building an outdoor staircase for his deck. He uses 8 screws for every 2 steps. If there are 14 steps, how many scre

4vir4ik [10]

Answer:

56 screws

Step-by-step explanation:

So, first we can find how many screws he needs.

So, if he uses 2 steps for 8 screws, and has 14 steps, that means he puts screws in 7 steps ( 14÷2)

then, if he uses 8 screws, each with 8 steps, we can do 8 times 7, which is 56 screws

3 0

2 years ago

What is the area of this trapezoid?​

What is the area of this trapezoid?