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tekilochka [14]
3 years ago
9

Does Cos (x+y)=cosx+cosy? Or not?

Mathematics
1 answer:
dangina [55]3 years ago
4 0
\bf \textit{Sum and Difference Identities}
\\ \quad \\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\ \quad \\
sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})
\\ \quad \\
\boxed{cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})}\impliedby notice
\\ \quad \\
cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}})
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[HELP ASAP 50 POINTS] 17, 18, 19, 20
Charra [1.4K]

17) f(x) = 16/(13-x).

In order to find domain, we need to set denominator expression equal to 0 and solve for x.

And that would be excluded value of domain.

13-x =0

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13-x +x = x.

13=x.

Therefore, domain is All real numbers except 13.



18).f(x) = (x-4)(x+9)/(x^2-1).

In order to find the vertical asymptote, set denominator equal to 0 and solve for x.

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Factoring out

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We have degrees of numberator and denominator are same.

Therefore, Horizontal asymptote is the fraction of leading coefficents.

That is 7/2.


20) f(x)=(x^2+3x-2)/(x-2).

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2>1.

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Therefore, there would no any Horizontal asymptote.

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