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.
N is greater than or equal to 5. (n <u>></u> 5)
Answer:
The pitcher paid $70.
The second baseman paid $63.
$70 -$63 = $7
Step-by-step explanation:
Answer: X = 9
Show Your Work:
All Angles & Sides = 60.
All Triangles = 180.
60 x 3 = 180.
Angle N:
7x - 3 = 60
7 * 9 - 3 = 60.
Answer:
4Joules
Step-by-step explanation:
According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
K is the elastic constant
e is the extension
If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'
can be gotten using k = f/e where
F = 6N, e = 3m
K = 6N/3m
K = 2N/m
Work done on an elastic string is calculated using 1/2ke².
If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;
1/2× 2× (2)²
= 4Joules
