1. Milky way 2. Sun 3. Moon 4. Earth 5. Stars

1, 4, 3, 2, 5

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2, 3, 4, 5, 1

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3, 4, 2, 5, 1

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4, 3, 2, 5, 1

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Solution

The correct option is D 3, 4, 2, 5, 1
The words given are different planetary bodies.
They must be arranged in the increasing or decreasing order of their sizes.
So for given words, the order will be:
Moon, Earth, Sun, Stars, MilkyWay
3,4,2,5,1
option C

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Similar questions

Find the mode from the given data.

1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2 , 3, 4, 1, 2, 3, 4, 1 ,2 , 3, 4, 5, 5, 1, 5, 5, 5, 3, 5, 1, 2, 4, 5, 6, 1, 2, 4, 2, 1

Solve:

[[5×{(−3)×(−4)}]÷{27 + (−14−3)}][[3×{(−4)×(−5)}]÷{27+(−13−4)}]
+[{27×(−2)}÷(−3−3)] + [{27×(−2)}÷(−3−3)]
+[{(−3)×(−3)×5}÷(−3)]]

$\int \frac{1}{2 + 3 s i n x} d x$

Q. Which fraction is smaller?

\frac{2}{10}, \frac{2}{7}

\frac{5}{7}, \frac{5}{9}

\frac{13}{14}, \frac{5}{6}

\frac{3}{13}, \frac{4}{14}

Find x, if

$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

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1. Milky way 2. Sun 3. Moon 4. Earth 5. Stars

The correct option is D 3, 4, 2, 5, 1The words given are different planetary bodies.They must be arranged in the increasing or decreasing order of their sizes.So for given words, the order will be:Moon, Earth, Sun, Stars, MilkyWay3,4,2,5,1option C

The correct option is D 3, 4, 2, 5, 1
The words given are different planetary bodies.
They must be arranged in the increasing or decreasing order of their sizes.
So for given words, the order will be:
Moon, Earth, Sun, Stars, MilkyWay
3,4,2,5,1
option C