Jigsaw,
No item 3. o correto é ...aresta lateral mede
2.5m
1) [tex3]1203_{10}=1000+200+0+3 \implies 1203=1\cdot10^3+2\cdot10^2+0\cdot10^1+3\cdot10^0[/tex3]
Logo, para determinar esse número na base 5, multiplique pelas potências de 5:
[tex3]1203_5=1\cdot5^3+2\cdot5^2+0\cdot5^1+3\cdot5^0\\
1203_5=1\cdot125+2\cdot25+0\cdot5+3\cdot1\\
1203_5=125+50+0+3\\
\boxed{1203_5=178_{(10)}}[/tex3]
2) [tex3]x^2+x-6 < 0 : Raízes: x = -3 \vee x=2\\
+++{\color{red} (-3)---(2)}+++ \implies\boxed{ -3 < x < 2}
[/tex3]
3)[tex3]Base ~quadrada: \\
d=4 = l_4\sqrt2 \implies l_4 =2\sqrt2\\
(\frac{5}{2})^2=h^2 +(\frac{d}{2})^2 \implies h^2 = \frac{25}{4} -4 \therefore h = \frac{3}{2}
\therefore V = \frac{S_b.h}{3}=\frac{(2\sqrt2)^2.3}{3.2}=\boxed{4m^3} [/tex3]
- fig1.jpg (6.53 KiB) Exibido 303 vezes
4) [tex3]1^o quadr. \implies cos \alpha> 0: sen\alpha > 0\\
(sen\alpha +cos \alpha)^2 = sen^2 \alpha +cos ^2\alpha +2sen \alpha cos \alpha = 1+ 2sen \alpha cos \alpha= 1+sen 2\alpha\\
sen \alpha + cos \alpha = \sqrt{\underbrace{1+sen2\alpha}_{>1}}\\
\therefore \boxed{sen \alpha + cos \alpha >1}
[/tex3]
