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2 years ago

Where does shearing often occur?

Geography

2 answers:

sweet [91]2 years ago

5 0

Answer: C.at a transform boundary

MakcuM [25]2 years ago

4 0

Answer:

transform boundary

Explanation:

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When Carla solved for the height from the ground to the point where the ladder

Westkost [7]

Answer:

See explanation

Explanation:

Given

Represent the vertical angle with $\theta$

$\theta = 75$

The question has incomplete details because the length of the ladder is not given; neither is the distance between the ladder and the wall given.

See attachment for illustration

So, this solution will be based on assumptions.

Represent

- The height from ground to the top of the ladder with y

- The length of the ladder with L

- The distance between the ladder and the wall with x

Carla could solve for y in any of the following ways:

1. Tan formula

$tan \theta = \frac{opp}{adj}$

In this case:

$tan \theta = \frac{x}{y}$

Multiply both sides by y

$y * tan \theta = \frac{x}{y} * y$

$y * tan \theta = x$

Divide both sides by tan

$y = \frac{x}{tan \theta}$

$y = \frac{x}{tan 75}$

This can be used if the distance (x) between the ladder and the wall is known.

Assume x = 15

$y = \frac{15}{tan 75}$

$y = 4.02$

2. Cosine formula

$cos \theta = \frac{adj}{hyp}$

In this case:

$cos \theta = \frac{y}{L}$

Multiply both sides by L

$L * cos \theta = \frac{y}{L} * L$

$Lcos \theta = y$

$y = Lcos \theta$

$y = Lcos75$

This can be used if the length (L) of the ladder is known.

Assume L = 15

$y = 15 * cos75$

$y = 3.88$

3 0

2 years ago

Goods from the city of Uruk have been found in various places around the Middle East. Which does this suggest?

Oksanka [162]

Answer:

trading was park of Uruks economy.

3 0

2 years ago

What is the distance between (0,-8) and (3,2)

DerKrebs [107]

Answer: $\Large\boxed{Distance=\sqrt{109}}$

Explanation:

Given information

(x₁, y₁) = (0, -8)

(x₂, y₂) = (3, 2)

Given the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute values into the given formula

$Distance=\sqrt{(3-0)^2+(2-(-8))^2}$

Simplify values in the parenthesis

$Distance=\sqrt{(3)^2+(2+8)^2}$

$Distance=\sqrt{(3)^2+(10)^2}$

Simplify the exponents

$Distance=\sqrt{9+100$

Simplify by addition

$\Large\boxed{Distance=\sqrt{109}}$

Hope this helps!! :)

Please let me know if you have any questions

6 0

1 year ago

QUESTION 10

Elodia [21]

Answer:

the process of improvement

4 0

3 years ago

How is the world changing? geography related

Hunter-Best [27]

Answer:

It may change structure do to human activity.

Explanation:

3 0

2 years ago