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.
Hey there!
We can solve systems two ways - substitution and elimination. It seems in this case that elimination is going to be easier.
Let's begin by multiplying the top equation by 3, and the bottom equation by 4:
3(5x - 4y = -24)
15x - 12y = -72
4(4x + 3y = 18)
16x + 12y = 72
Next, we'll add the two equations together to get rid of, or eliminate, the y variable and solve for the x variable:
(15x - 12y = -72) + (16x + 12y = 72)
31x = 0
x = 0
Next, we'll plug the value of the x variable into the top equation and solve for the y variable:
5(0) - 4y = -24
-4y = -24
y = 6
Therefore, x = 0 and y = 6.
Hope this helps!! :)
Answer:
X= 5, -9
Step-by-step explanation:
(X+2)² =49
Expand the bracket,
(X+2) (X+2)= 49
Apply the distributive property;
X(X+2) +2 (X+2) =49
X²+2X+2X+4=49
X²+4X+4=49
Move 49 to the left side of the equation;
X²+4X+4-49=0
X²+4X-45=0
Apply Factorisation method;
Consider the form
a²+bx+c=0
Find two numbers whose sum is equal to b and whose product is equal to c.
Comparing with our equation;
B =4 and C =45
We can use 9 and -5, this is because ;
9+(-5)=4 and 9*(-5)= 45.
Replace X +4X-45=0 with (X-5) (X+9)=0
Therefore;
X-5=0
X+9=0
Moving to the left side of the equation;
Therefore X= 5 and -9
Answer:
2/15
Step-by-step explanation:
1/5 x 2/3 = 2/15 which can't be simplified so 2/15 is the correct answer
