Answer:
Naomi had the correct measurement the ring weighed 0.625 carats
Step-by-step explanation:
* The diamond ring weighed 5/8 carats
- To know who is right lets change the fraction 5/8 to a decimal number
∵ 5/8 means 5 ÷ 8
- 5 is smaller than 8 so we will multiply it by 10 and
insert decimal point in the quotient (answer of division)
∴ It will be 50 ÷ 8 = 6 and remainder 2/8 ⇒ 1/4
∴ The quotient = 0.6 and remainder 1/4
- 1 is smaller than 4 so we will multiply it by 10
∴ It will be 10 ÷ 4 = 2 and remainder 2/4 ⇒ 1/2
∴ The quotient = 0.62 and remainder 1/2
- 1 is smaller than 2 so we will multiply it by 10
∴ It will be 10 ÷ 2 = 5 without remainder
∴ The quotient = 0.625
* Naomi had the correct measurement the ring weighed 0.625 carats
Answer:
The area of the shaded region is 
Step-by-step explanation:
we know that
The area of the shaded region is equal to the area of the rectangle minus the area of the circle
The rectangle is a square
so
we have
assume
substitute the values
Answer:
There are 31 students in the class.
Step-by-step explanation:
![\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]](https://tex.z-dn.net/?f=%5Crightarrow%20z%5E4%3D-625%5C%5C%5C%5C%5Crightarrow%20z%3D%28-625%2B0i%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5C%5C%5C%5C%5Crightarrow%20x%2Biy%3D%28-625%2B0i%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5C%5C%5C%5C%20x%3Dr%20%5Ccos%20A%5C%5C%5C%5Cy%3Dr%20%5Csin%20A%5C%5C%5C%5Cr%20%5Ccos%20A%3D-625%5C%5C%5C%5C%20r%20%5Csin%20A%3D0%5C%5C%5C%5Cx%5E2%2By%5E2%3D625%5E%7B2%7D%5C%5C%5C%5Cr%5E2%3D625%5E%7B2%7D%5C%5C%5C%5C%7Cr%7C%3D625%5C%5C%5C%5C%20%5Ctan%20A%3D%5Cfrac%7B0%7D%7B-625%7D%5C%5C%5C%5C%20%5Ctan%20A%3D0%5C%5C%5C%5C%20A%3D%5Cpi%5C%5C%5C%5C%5Crightarrow%20z%3D%20%5B625%28%5Ccos%20%282k%20%5Cpi%2Bpi%29%20%2Bi%20%5Csin%20%282k%5Cpi%2B%20%5Cpi%29%5D%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5C%5C%5C%5Ck%3D0%2C1%2C2%2C3%2C4%2C....%5C%5C%5C%5C%5Crightarrow%20z%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7B%282k%20%5Cpi%2Bpi%29%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B%282k%5Cpi%2B%20%5Cpi%29%7D%7B4%7D%5D%20)
![\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]](https://tex.z-dn.net/?f=%5Crightarrow%20z_%7B0%7D%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7Bpi%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B%5Cpi%29%7D%7B4%7D%5D%5C%5C%5C%5C%5Crightarrow%20z_%7B1%7D%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7B3%5Cpi%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B4%7D%5D%5C%5C%5C%5C%20%5Crightarrow%20z_%7B2%7D%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%5D%5C%5C%5C%5C%20%5Crightarrow%20z_%7B3%7D%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7B7%5Cpi%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B7%5Cpi%7D%7B4%7D%5D)
Argument of Complex number
Z=x+iy , is given by
If, x>0, y>0, Angle lies in first Quadrant.
If, x<0, y>0, Angle lies in Second Quadrant.
If, x<0, y<0, Angle lies in third Quadrant.
If, x>0, y<0, Angle lies in fourth Quadrant.
We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is
![\rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]](https://tex.z-dn.net/?f=%20%5Crightarrow%20z_%7B2%7D%3D%28625%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5B%5Ccos%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%20%2Bi%20%5Csin%20%5Cfrac%7B5%5Cpi%7D%7B4%7D%5D)
What you first need to do is make an equation. Then plug that equation into photomath and you get your answer
