Answer:
Fourth degree polynomial (aka: quartic)
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Work Shown:
There isnt much work to show here because we can use the fundamental theorem of algebra. The fundamental theorem of algebra states that the number of roots is directly equal to the degree. So if we have 4 roots, then the degree is 4. This is assuming that there are no complex or imaginary roots.
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If you want to show more work, then you would effectively expand out the polynomial
(x-m)(x-n)(x-p)(x-q)
where
m = 4, n = 2, p = sqrt(2), q = -sqrt(2)
are the four roots in question
(x-m)(x-n)(x-p)(x-q)
(x-4)(x-2)(x-sqrt(2))(x-(-sqrt(2)))
(x-4)(x-2)(x-sqrt(2))(x+sqrt(2))
(x^2-6x+8)(x^2 - 2)
(x^2-2)(x^2-6x+8)
x^2(x^2-6x+8) - 2(x^2-6x+8)
x^4-6x^3+8x^2 - 2x^2 + 12x - 16
x^4 - 6x^3 + 6x^2 + 12x - 16
We end up with a 4th degree polynomial since the largest exponent is 4.
2.7 dollars. You round to the nearest tenth which means the first number after the decimal, which is six. Since there is a 5 after six, and 5 can be rounded positively, so the 6 becomes a seven
SAS, side angle side congruency
Answer:
id take it but don know how
Step-by-step explanation:
