Answer:
Attachment 1 : Option C
Attachment 2 : Option A
Step-by-step explanation:
( 1 ) Expressing the product of z1 and z2 would be as follows,
![14\left[\cos \left(\frac{\pi \:}{5}\right)+i\sin \left(\frac{\pi \:\:}{5}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{3\pi \:}{2}\right)+i\sin \left(\frac{3\pi \:\:}{2}\right)\right]](https://tex.z-dn.net/?f=14%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%20%5C%3A%7D%7B5%7D%5Cright%29%2Bi%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%5C%3A%5C%3A%7D%7B5%7D%5Cright%29%5Cright%5D%5Ccdot%20%5C%3A2%5Csqrt%7B2%7D%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B3%5Cpi%20%5C%3A%7D%7B2%7D%5Cright%29%2Bi%5Csin%20%5Cleft%28%5Cfrac%7B3%5Cpi%20%5C%3A%5C%3A%7D%7B2%7D%5Cright%29%5Cright%5D)
Now to solve such problems, you will need to know what cos(π / 5) is, sin(π / 5) etc. If you don't know their exact value, I would recommend you use a calculator,
cos(π / 5) = 
,
sin(π / 5) = 
cos(3π / 2) = 0,
sin(3π / 2) = - 1
Let's substitute those values in our expression,
![14\left[\frac{\sqrt{5}+1}{4}+i\frac{\sqrt{2}\sqrt{5-\sqrt{5}}}{4}\right]\cdot \:2\sqrt{2}\left[0-i\right]](https://tex.z-dn.net/?f=14%5Cleft%5B%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B4%7D%2Bi%5Cfrac%7B%5Csqrt%7B2%7D%5Csqrt%7B5-%5Csqrt%7B5%7D%7D%7D%7B4%7D%5Cright%5D%5Ccdot%20%5C%3A2%5Csqrt%7B2%7D%5Cleft%5B0-i%5Cright%5D)
And now simplify the expression,
The exact value of 
= 
and 
= 
Therefore we have the expression 
, which is close to option c. As you can see they approximated the solution.
( 2 ) Here we will apply the following trivial identities,
cos(π / 3) = 
,
sin(π / 3) = 
,
cos(- π / 6) = ,
sin(- π / 6) = 
Substitute into the following expression, representing the quotient of the given values of z1 and z2,
![15\left[cos\left(\frac{\pi \:}{3}\right)+isin\left(\frac{\pi \:\:}{3}\right)\right] \div \:3\sqrt{2}\left[cos\left(\frac{-\pi \:}{6}\right)+isin\left(\frac{-\pi \:\:}{6}\right)\right]](https://tex.z-dn.net/?f=15%5Cleft%5Bcos%5Cleft%28%5Cfrac%7B%5Cpi%20%5C%3A%7D%7B3%7D%5Cright%29%2Bisin%5Cleft%28%5Cfrac%7B%5Cpi%20%5C%3A%5C%3A%7D%7B3%7D%5Cright%29%5Cright%5D%20%5Cdiv%20%5C%3A3%5Csqrt%7B2%7D%5Cleft%5Bcos%5Cleft%28%5Cfrac%7B-%5Cpi%20%5C%3A%7D%7B6%7D%5Cright%29%2Bisin%5Cleft%28%5Cfrac%7B-%5Cpi%20%5C%3A%5C%3A%7D%7B6%7D%5Cright%29%5Cright%5D)
⇒
![15\left[\frac{1}{2}+\frac{\sqrt{3}}{2}\right]\div \:3\sqrt{2}\left[\frac{\sqrt{3}}{2}+-\frac{1}{2}\right]](https://tex.z-dn.net/?f=15%5Cleft%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Cright%5D%5Cdiv%20%20%5C%3A3%5Csqrt%7B2%7D%5Cleft%5B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2B-%5Cfrac%7B1%7D%7B2%7D%5Cright%5D)
The simplified expression will be the following,
or in other words 
or 
The solution will be option a, as you can see.
Answer:
3rd one is right answers so just put that only
Answer: 3/8
Step-by-step explanation:
Since it is a fair coin, then generally, P(Head) = P(Tail) = ½
And since we've been asked to find the probability that the number of heads in the first two tosses be equal to the number of heads in the second two tosses, tossing a fair coin four times, the possible outcomes of having equal number of heads in first two tosses and second two tosses becomes:
[HHHH] or [HTHT] or [THTH] or [TTTT] or [HTTH] or [THHT]
=[½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½]
=1/16 * 6
=6/16
=3/8.
There is no attachment to your phone question. I look forward to answer your question.
Answer:
Attached is your answer
Step-by-step explanation:
