Jigsaw,
2.1) Usando o T. raizes racionais:
Divisores de -2(p): 1,2,-1,-2
Divisores de 6(q), 6,2,3,1, -1,-2,-3,-6
Possíveis raízes inteiras: [tex3]\pm 1, \pm2, [/tex3]
Testando encontraremos como raiz [tex3]\boxed{x=-2}[/tex3]
Reduzindo para equação de 2
o usando Briott Ruffini teremos
o grau: (x+2)(6x
2-x-1)=0
[tex3]\Delta = 25 \implies x = \frac{1\pm5}{12} \therefore x = \frac{1}{2} \vee \boxed{x = -\frac{1}{3}}[/tex3]
2.2)
- fig1.jpg (23.48 KiB) Exibido 219 vezes
2.3) [tex3]V_{cil}=\pi.r^2.h=\pi .2^2.8 =32 \pi\\
V_{con} = \frac{S_b.8}{3} \\
S_b = \frac{{l^2\sqrt3}}{4}\\
l_\triangle=r\sqrt3 = 2\sqrt3 \implies S_b=\frac{12.\sqrt3}{4}=3\sqrt3\\
\therefore V_{con} = 8\sqrt3\\
V=V_{cil}-V_{con} = \boxed{32\pi-3\sqrt3}
[/tex3]
2.4) [tex3]sn^2a +cos^2 a =1 \implies (\frac{3}{5})^2+cos^2a=1 \therefore cos a =\frac{4}{5}\implies tg(a) =\frac{3}{4} \\
sen^2b+cos^2b=1 \implies sen^2b+(\frac{4}{7})^2=1 \therefore senb = \frac{\sqrt{33}}{7}\implies tg(b) = \frac{\sqrt{33}}{4}\\
\tg(a+b) = \frac{tg(a)+tg(b)}{1+tg(a).tg(b)} = \frac{\frac{3+\sqrt{33}}{4}}{1+\frac{3\sqrt{33}}{16}}= \frac{\frac{3+\sqrt{33}}{4}}{\frac{16+3\sqrt{33}}{16}}=\\
\boxed{\frac{12+4\sqrt{33}}{16+3\sqrt{33}}} [/tex3]