Answer:
i think the answer is 1000 kg but i might be wrong
I assume you meant to say

Given that <em>x</em> = √3 and <em>x</em> = -√3 are roots of <em>f(x)</em>, this means that both <em>x</em> - √3 and <em>x</em> + √3, and hence their product <em>x</em> ² - 3, divides <em>f(x)</em> exactly and leaves no remainder.
Carry out the division:

To compute the quotient:
* 2<em>x</em> ⁴ = 2<em>x</em> ² • <em>x</em> ², and 2<em>x</em> ² (<em>x</em> ² - 3) = 2<em>x</em> ⁴ - 6<em>x</em> ²
Subtract this from the numerator to get a first remainder of
(2<em>x</em> ⁴ + 3<em>x</em> ³ - 5<em>x</em> ² - 9<em>x</em> - 3) - (2<em>x</em> ⁴ - 6<em>x</em> ²) = 3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3
* 3<em>x</em> ³ = 3<em>x</em> • <em>x</em> ², and 3<em>x</em> (<em>x</em> ² - 3) = 3<em>x</em> ³ - 9<em>x</em>
Subtract this from the remainder to get a new remainder of
(3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3) - (3<em>x</em> ³ - 9<em>x</em>) = <em>x</em> ² - 3
This last remainder is exactly divisible by <em>x</em> ² - 3, so we're left with 1. Putting everything together gives us the quotient,
2<em>x </em>² + 3<em>x</em> + 1
Factoring this result is easy:
2<em>x</em> ² + 3<em>x</em> + 1 = (2<em>x</em> + 1) (<em>x</em> + 1)
which has roots at <em>x</em> = -1/2 and <em>x</em> = -1, and these re the remaining zeroes of <em>f(x)</em>.
Energy is absorbed by the atom.
Emission lines occur only in the case of transitions from higher to lower energy levels in which energy is emitted.
Answer:
The minimum feature size is 136.6 km.
Explanation:
It is given that,
Distance from planet Mars and Earth is, 
Diameter of the objective lens, d = 0.977 m
We need to find the minimum feature size, on the surface of Mars that your telescope can resolve for you. The expression for resolving distance is given by :

So, the minimum feature size is 136.6 km.
Answer:
The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.
Explanation:
Momentum before = momentum after
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂
10 m/s = 5 v₁ + 4 v₂
Assuming an elastic collision, kinetic energy is conserved.
½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²
m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²
(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²
280 m²/s² = 5 v₁² + 4 v₂²
Substituting:
v₂ = (10 − 5 v₁) / 4
280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²
280 = 5 v₁² + (10 − 5 v₁)² / 4
1120 = 20 v₁² + (10 − 5 v₁)²
1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²
0 = 45 v₁² − 100 v₁ − 1020
0 = 9 v₁² − 20 v₁ − 204
0 = (9 v₁ + 34) (v₁ − 6)
v₁ = -3.78 m/s or 6 m/s
u₁ = 6 m/s, so v₁ = -3.78 m/s. Solving for v₂:
v₂ = (10 − 5 v₁) / 4
v₂ = 7.22 m/s
The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.