What would happen to the propagation of a neural impulse if the membrane potential (voltage) at the synaptic terminal did not cross the threshold to open voltage-gated calcium channels

Hello the options related to your question is missing attached below are the missing options

answer : Calcium channels would remain closed, causing the neural impulse to travel back to the cell body. ( B )

Explanation:

When the membrane potential at the synaptic terminal does not cross the given threshold to an open Voltage-gated calcium channels, the propagation of the neural impulse will travel back to the cell body and this is caused by the calcium channels to remaining  closed

9.9 years

a = p(1 + rt) . . account balance after time t at rate r starting with principal p

3000 = 2500(1 + 0.0202t) . . filling in the given numbers

1.2 = 1 + 0.0202t . . divide by 2500

0.2 = 0.0202t . . . subtract 1

0.2/0.0202 = t ≈ 9.901

it will take about 9.9 years for the account balance to reach $3000.

it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.

explanation:

the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.

countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.