Ya I believe that is the answer. I got the exact same thing.
2m⁴ - 18n⁶
2(m⁴) - 2(9n⁶)
2(m⁴ - 9n⁶)
2(m⁴ - 3m²n³ + 3m²n³ - 9n⁶)
2[m²(m²) - m²(3n³) + 3n³(m²) - 3n³(3n³)]
2[m²(m² - 3n³) + 3n³(m² - 3n³)]
2(m² + 3n³)(m² - 3n³)
Yes, substitute a=2.
14 - 4 + 6 - 6 = -2
8 - 10 = -2
So distance = rate × time. You would do 3/4 × 5.5, which would get you 4.125.
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
