222+203 is rounded up to what?

1 answer:

3 0

$222: 220 \\ 203 = 200 \\ \\ 222 + 203 = 425 \\ 220 + 200 = 420 \\ \\ \\ \\ Good \\ luck \\ on \\ your \\ assignment \\ \\ enjoy \\ your \\ day \\ \\ \\ MeIsKaitlyn :)$

$Remember$ ↓

$1-5$ would rounded $downward$
$6-9$ would be rounded $upward$

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Harlamova29_29 [7]

Answer:

$\huge\boxed{\sf Option \ 2}$

Step-by-step explanation:

$\sf (\frac{4}{5})^3 \\\\$

This means that 4/5 repeats itself 3 times.

So,

$\sf = \frac{4}{5} * \frac{4}{5} * \frac{4}{5} \\\\= \frac{64}{125} \\\\\rule[225]{225}{2}$

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3 years ago

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Dvinal [7]

Its ‘B’

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Bezzdna [24]

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Find the solution of the following equation whose argument is strictly between 270^\circ270 ∘ 270, degree and 360^\circ360 ∘ 360

Natasha2012 [34]

$\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]$

$\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]$