Your answer is 2.50 (And the rest of the 0s, but they aren't needed...) Since that is greater then 1, but less than 10.
Answer: 90 + 4, third choice and 90 - -4, last choice
Her score is obviously higher when the mistake is fixed, adding the 4 points gives 94.
Answer: Area of square PQRS is 81cm^2 and of ABCD is 25cm^2
You know its a square. Which means all the sides are equal and same length. If you dont know what's a square then here is the definition:
Square is a plane figure with four equal sides and four right (90°) angles.
So now we know both of are squares, in case of the first square, PQRS, one side length is already given which is PQ=9cm. Now one side is 9cm which means all the sides are 9cm's (PQ=9cm, QR=9cm, RS=9cm,SP=9cm). So to find area just multiply them:
9cm * 9cm = 81cm^2
Now you found the area of the first square. So for the second square, ABCD, one side length is 5 cm so all the rest three sides are also 5 cm. Again, multiply them in order to find the area:
5cm * 5cm = 25cm^2
So square PQRS is 81cm^2 to that of square ABCD is 25cm^2
CAUTIONS: IT IS NOT THAT HARD YA KNOW
By the way, hope it helps! ^^
![\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)
Answer:
71
Step-by-step explanation:
[6x4(15/5)]+[2^2+(1x-5)]
BIDMAS (aka PEDMAS etc.)
Brackets
Indices
Division/Multiplication
Addition/Subtraction
[6 x 4(15/5)] + [2^2 + (1x-5)]
[6 x 4(3)] + [2^2 + (-5)]
[6 x 4(3)] + [4 + -5]
[6 x 12] + [4 + -5]
[72] + [4 + -5]
72 + [-1]
72 + - 1
72 - 1
71
