There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4) where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a±bi [±=+/-]
So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
=x^4+12x^3+77x^2+196x+164 [if you decide to expand]
Answer: The relation is a function.
Step-by-step explanation: Since there is one value of y for every value of x in (-2,4), (3,7), (0,8), (5,8), and (1,6), this relation is a function.
I hope this helps you out!
substitution and elimination means that you need to use one equation and substitute it in place of some variable in the other equation.
Consider the above two equations.
Take equation 1, which is 4x-2y=22 => x=(22+2y)/4
substitute the x value gained above in equation 2.
so, 2((22+2y)/4)+4y=6
22+2y+8y=12 => 10y = -10 => y= -1.
Substitute y= -1 in x value obtained in the beginning.
So, x= (22 - 2)/4 => 5.
There fore, x= 5 and y= -1
Hope it helps.
I answered this but i couldn’t take multiple pictures. however i found a website with the same answer as me after i answered your question. here : https://www.quora.com/How-do-I-integrate-Integrate-sinxdx-sin-3x+cos-3x
Answer:
I'm gemini
Step-by-step explanation:
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