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Bad White [126]
2 years ago
5

QUESTION 33.1 Round the following number up to the nearest whole numberR13492.39​

Mathematics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer:

the answer is R13492

Step-by-step explanation:

hope it helps

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A(R+T)=W solve for T
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(A×R)+(A×T)=W
AR+AT=W
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AT=W-AR
divide both sides by A
T=(W-AR)/A
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Sam received a 15 percent tip for delivering pizzas. If the cost of the pizzas was $247.62, what was the total amount he collect
saveliy_v [14]
247.62 x 0.15 = 37.143

round to $37.14

I hope that helped :)
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A supermarket randomly gives each shopper a free two liter bottle of cola during the Saturday shopping hours. The supermarket se
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3 years ago
Find the solution of the following equation whose argument is strictly between 270^\circ270 ∘ 270, degree and 360^\circ360 ∘ 360
Natasha2012 [34]

\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}]

\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}]

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}]

5 0
3 years ago
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