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.
Reorder the terms
112 + 3x = 100 + 5x
112 + 3x + (-5x) = 100 + 5x + (-5x)
Combine the like terms
5x + (-5x) = 0
112 + (-2x) = 100 + 0
112 + (-2x) = 100
112 + (-112) + (-2x) = 100 + (-112)
Combine like terms again
112 + (-112) =0
0 + (-2x) = 100 + (-112)
100 + (-112) = (-112) = (-12)
-2x = -12
-12 / -2 = 6
x = 6 <span>✓</span>
Check your work
3 * 6 + 112 = 130
5 * 6 + 110 = 130
130 = 130 ✓
Answer:
5 x + 3 = 23
Step-by-step explanation:
I think it should be 1.7?
a^2 + b^2 = c^2
c^2 = 16
Square root of 16 is 4.
1 + 3 = 4
Square root of 1 is 1 and square root of 3 is given, which is 1.7.
An equilateral has two right triangles out together, so basically that would be a rectangle.
1 x 1.7 = 1.7
Yeah y=9
hope this helped even tho I’m kinda late haha
