Can polynomials have odd roots (like f(x)= cube root of x)?? Fast!! I’ll give branliest!
2 answers:
I think you're asking if it's possible to have a cube root, fifth root, 7th root, etc of a number as a solution to f(x). The answer is yes it's possible.
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Example:
f(x) = x^3 - 29
This function has one real-number root of (cube root of 29) and the other two roots are complex or imaginary roots.
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Answer:
$403.15
Step-by-step explanation:
Principal loan amount is the total amount minus down-payment:
Knowing that , the monthly payments can be calculated using the formula:
Hence, the monthly payment is $403.15
Answer: a) degree and sign
b) end behavior: left side → +∞, right side → -∞
c) x-intercepts: x = -1.3, 0.3, 1.0
<u>Step-by-step explanation:</u>
end behavior can be determined by two things:
1) the degree of the polynomial:
- if the degree is an even number, then the end behavior will be the same for both the left and right sides.
- if the degree is an odd number, then the end behavior will be different for both the left and right sides.
2) the sign of the leading coefficient:
- If the leading coefficient is positive, then the end behavior of the right side goes to positive infinity
- If the leading coefficient is negative, then the end behavior of the right side goes to negative infinity
W(x) = -5x³ + 7x - 2
Degree: 3 (odd)
Leading Coefficient: negative
So, end behavior is: right side goes to negative infinity, right side goes to positive infinity.
See attachment for x-intercepts. <em>I set the x-axis to represent tenths </em>
