Answer:
1. dendrite = directs impulses toward the soma.
2. axon = conducts impulses toward the synaptic terminal.
3. perikaryon = region surrounding nucleus.
4. collateral branches = main branches of an axon.
5. synaptic terminal = enlarged end of an axon.
6. synaptic vesicles = contains neurotransmitters.
7. axon hillock = connects the cell body and axon.
8. Nissl bodies = clusters of RER and free ribosomes.
9. telodendria = fine branches of an axon.
10. myelinated internode = part of axon covered by Schwann cell.
11. neurilemma = Schwann cell's plasma membrane.
12. axolemma = membrane of the axon.
13. astrocyte = Forms the blood-brain barrier.
14. cell body = soma.
Answer:$24 for it all
Explanation:
Honestly no chance of getting a free delivery unless the place you ordering from has a "order over (a certain amount) get it free delivery"
Answer:
I think number 5. Because for a toddler climbing up and down of a chair is very dangerous so I think for the nurse she should plan safe methods to improve the toddler's safety at home and there could be something dangerous and harmful in those boxes
The consequences if presynaptic action potentials In an axon release insufficient acetylcholine to depolarize a skeletal muscle fiber to threshold
Explanation:
When an action potential reaches a neuromuscular junction, it causes acetylcholine to be released into this synapse. The acetylcholine binds to the nicotinic receptors concentrated on the motor end plate, a specialized area of the muscle fibre's post-synaptic membrane.
Acetylcholine is the neurotransmitter used at the neuromuscular junction—in other words, it is the chemical that motor neurons of the nervous system release in order to activate muscles. ... In the brain, acetylcholine functions as a neurotransmitter and as a neuromodulator.
Answer:
You don't, because it's false. If all black dots happen to be on the line y=0 and white dots on the line y=π (and the rest of the plane is neither white nor black), there is no such pair.
Now if each point of the plane were either black or white (and there were infinitely many of each type), that would be different. In fact, it is sufficient to have at least one of each color.
Why? Pick any two points A and B that have different colors. Starting at A , we can reach B using a finite number of steps, each of length exactly 1: just go directly towards B until the distance becomes less than 1, and at the end, if we didn't reach B exactly, we make two steps "to the side and back" to reach it. (Formally, if you are currently at C , imagine circles with radius 1 centered at B and C . Pick one of their two intersections, go from C to that intersection and from there to B .)
As the first and the last point on this path have opposite colors, there has to be a pair of consecutive points with opposite colors, q.e.d.
(Alternately, you could prove the new statement by contradiction. Pick any black point. All points in distance 1 from that point have to be black. This is the circle with radius 1. All points in distance 1 from those points have to be black as well. Here we can observe that the set of all points known to be black at this moment is the entire disc of radius 2 centered where we started. Continuing this argument, we can now grow the black disc indefinitely and thus prove that the entire plane has to be black, which is the contradiction we seek. Of course, this is basically the same proof as above, just seen from a different point of view.)
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