Db is a diameter of circle o. if macb = 30°, what is the mdba?

30°

50°

60°

70°

90°

m∠ABD = 60°

Step-by-step explanation:

From the given figure :

Given : m∠ACB = 30°

To find : m∠DBA

Solution : Since, the angle formed in the same segment are equal in the circle

⇒ m∠ACB = m∠ADB

Since, m∠ACB = 30°

⇒ m∠ADB = 30°

Now, Angle formed in the semicircle is right angle

⇒ m∠BAD = 90°

So, by using angle sum property of a triangle in the ΔADB. We get,

m∠ADB + m∠ABD + m∠BAD = 180

⇒ 30 + m∠ABD + 90 = 180

⇒ m∠ABD + 120 = 180

⇒ m∠ABD = 60°

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

step-by-step explanation:

*calories burned = (weight)(calorie factor)*

calories burned = 180 (0.65)

calories burned = 117

the calories he used up is 117, which is 65% of his weight.