Brad is correct.
Polynomials do not include negative exponents; this means there can be no variables in the denominator. If two polynomials do not divide evenly, the remainder is written as a fraction with the divisor (which can include a variable) in the denominator. This is not a polynomial.
The expressions with radicals which are variables and numbers raised to a fractional indices are simplified as follows.
13. √(9·x) = 3·√x
14. √(4·y) = 2·√y
15. √(8·x²) = 2·x·√2
16. √(9·x²) = 3·x
17. √(3·x²) = x·√3
18. √(5·y²) = y·√5
19. √(13·x²) = x·√(13)
20. √(29·y²) = y·√(29)
21. √(64·y²) = 8·y
22. √(125·a²) = 5·a·√5
23. ∛(16) = 2·∛2
24. √(50·a²·b) = 5·a·√(2·b)
<h3>What are radicals expressions?</h3>
A radical expression is one that contains the radical (square root or nth root) sign, √.
13. √(9·x)
√(9·x) = √(3²·x) = 3·√x
14. √(4·y)
√(4·y) = √(2²·y) = 2·√y
15. √(8·x²)
√(8·x²) = √(4 × 2·x²) = √(2² × 2·x²)
√(2² × 2·x²) = √(2²·x² × 2) = 2·x·√2
16. √(9·x²)
√(9·x²) = √(3²·x²) = 3·x
17. √(3·x²)
18. √(5·y²)
√5 × √(y²) = √5 × y = y·√5
19. √(13·x²)
√(13·x²) = √(13) × √x² = √(13) × x = x·√(13)
20. √(29·y²)
√(29·y²) = √(29) × √(y²) = √(29) × y = y·√(29)
21. √(64·y²)
√(64·y²) = √(8²·y²) = √(8²) × √(y²) = 8 × y = 8·y
22. √(125·a²)
√(125·a²) = √(25 × 5 × a²) = √(25) × √5 × √(a²) = 5 × √5 × a
5 × √5 × a = 5·a·√5
23. ∛(16)
∛(16) = ∛(16) = ∛(8 × 2) = ∛(2³ × 2) = 2·∛2
24. √(50·a²·b)
√(50·a²·b) = √(25 × 2 × a² × b) = √(5² × 2 × a² × b) = √(5² × a² × 2 × b)
√((5² × a²) × 2 × b) = 5·a·√(2·b)
Learn more about simplifying expressions with radicals here:
brainly.com/question/13114751
#SPJ1
Answer:
$16,700
Step-by-step explanation:
Subtract: 38428
<u>-21728</u>
16700
You will need to add 10 to the hundred's place by subtracting 1 from the thousandth's place: 37,4(+10)28. Then you subtract as usual.
I want to say it is A because if you simplify it then it should say 36x/6x. I hope this helps! If wrong correct me..
Answer:
The answer is "The discriminant is positive"
Step-by-step explanation:
When the discriminant is positive there is two real solutions.
When the discriminant is equal to 0, there is one real solution.
When the discriminant is negative, there is no real number solutions.
