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Neptune's distance from the sun is about 2.8 x 10^9 miles.
Earths distance from the sun is about 30 times shorter.
What is Earths approximate distance, in miles, from the sun?

Explanation:

x = distance from sun to earth

y = distance from sun to neptune = 2.8 * 10^9 miles

"earths distance from the sun is about 30 times shorter" compared to neptune's distance, so,

y = 30x

x = y/30

x = (2.8 * 10^9)/30

x = (2.8 * 10^9)/(3.0 * 10^1)

x = (2.8/3.0) * (10^9/10^1)

x = 0.933333333333333 * 10^(9-1)  ... which is approximate

x = 0.933333333333333 * 10^8

x = 0.93 * 10^8 see note1 below

x = (9.3 * 10^(-1)) * 10^8 ... see note 2 below

x = 9.3 * (10^(-1)*10^8)

x = 9.3 * 10^(-1+8)

x = 9.3 * 10^7

Notes:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this