For every 3.3 km rock is buried, the pressure placed upon it increased 0.1 gigapascals. if rock undergoing metamorphosis is buried at 16 km what is the pressure placed on that rock?

For every 3.3 km rock is buried, the pressure placed upon it increased 0.1 Giga pascals.

We need to find the pressure placed on that rock at 16 km.

Let us divide 16 km by 3.3 first to find the number of 0.1 Giga pascals would be there.

16/3.3.

On dividing 16 by 3.3, we get 4.8485.

In order to find the number the pressure placed on that rock, we need to multiply 0.1 by 4.8485, we get

0.48485 Giga pascals.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

there still no table