You can't prove that generalization is likely correct. <em />It is told that generalization is never correct and some has told that generalization is sometimes correct, so I guess that would depend on your thoughts, opinions.
hope this helped!!
~Melany~ ;)

5 0

4 years ago

What exactly are prime numbers?

nignag [31]

Prime numbers are numbers that only have a factor of one and its self.

4 0

3 years ago

Read 2 more answers

Find a possible formula for a fourth degree polynomial g that has a double zero at -2, g(4) = 0, g(3) = 0, and g(0) = 12. g(x) =

astraxan [27]

<h2>Answer:</h2>

The possible formula for a fourth degree polynomial g is:

$g(x)=\dfrac{1}{4}(x^4-3x^3-12x^2+20x+48)$

<h2>Step-by-step explanation:</h2>

We know that if a polynomial has zeros as a,b,c and d then the possible polynomial form is given by:

$f(x)=m(x-a)(x-b)(x-c)(x-d)$

Here the polynomial g has a double zero at -2, g(4) = 0, g(3) = 0.

This means that the polynomial g(x) is given by:

$g(x)=m(x-(-2))^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x+2)^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x^2+2^2+2\times 2\times x)(x(x-3)-4(x-3))\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-3x-4x+12)\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-7x+12)\\\\i.e.\\\\g(x)=m[x^2(x^2-7x+12)+4(x^2-7x+12)+4x(x^2-7x+12)]\\\\i.e.\\\\g(x)=m[x^4-7x^3+12x^2+4x^2-28x+48+4x^3-28x^2+48x]\\\\i.e.\\\\g(x)=m[x^4-7x^3+4x^3+12x^2+4x^2-28x^2-28x+48x+48]\\\\i.e.\\\\g(x)=m[x^4-3x^3-12x^2+20x+48]$

Also,

$g(0)=12$

i.e.

$48m=12\\\\i.e.\\\\m=\dfrac{12}{48}\\\\i.e.\\\\m=\dfrac{1}{4}$

Hence, the polynomial g(x) is given by:

$g(x)=\dfrac{1}{4}(x^4-3x^3-12x^2+20x+48)$

8 0

3 years ago