3x^2y^3+5x^3y^3-6+x^3y^2-2
= (3x^2y^3+5x^3y^3+x^3y^2)-(6+2)
= x^2y^2(3y+5xy+x)-8.
The final answer is x^2y^2(3y+5xy+x)-8~
Answer:
- zeros are {-2, 3, 7} as verified by graphing
- end behavior: f(x) tends toward infinity with the same sign as x
Step-by-step explanation:
A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.
<h3>Zeros</h3>
The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.
<h3>End behavior</h3>
The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.
- x → -∞; f(x) → -∞
- x → ∞; f(x) → ∞
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<em>Additional comment</em>
The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.
We know the x^2 coefficient is the opposite of the sum of the zeros:
-(7 +(-2) +3) = -8 . . . . x^2 coefficient
And we know the constant is the opposite of the product of the zeros:
-(7)(-2)(3) = 42 . . . . . constant
These checks lend further confidence that the zeros are those given.
(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)
Yes, the price is proportional because you can reduce their fractions:
16/4 = 4
24/6 = 4
36/9 = 4
56/14 = 4
Suppose the integers are n , n+2 , n+4 and n+6.
84=n+(n+2)+(n+4)+(n+6)=4n+12.
Subtract 12 from both ends to get.
72=4n.
Divide both ends by 4 to get.
n=18.
So the integers are: 18 , 20 , 22 , 24.
I think the answer is 21. I’m not sure but if I remember it’s 40 + 20 = 60. 180 - 60 = 120 and then it’s a vertical angle which is equal so then it would be 120 + 39 = 159. Then 180 - 159 = 21. Sorry if this is wrong I haven’t done this in a while. Have a good day!