Cos^-1 sqrt 2/2 exact value

Mathematics

1 answer:

Leto [7]3 years ago

8 0

$\sf cos^-1 (\frac{\sqrt{2}}{2})=cos^-1 (\frac{1}{ \sqrt{2}})= \frac{ \pi }{4} ={\boxed{45\°}$

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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) =

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<h2>Answer:</h2>

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<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:

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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:

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