Question

help ofer toate punctele ​

Answer 1

$1.~~a_4=-4,~a_8=12 \\\\ a_4=-4 \Rightarrow a_1+3r=-4\\\\a_8=12 \Rightarrow a_1+7r=12\\\\\left\{\begin{array}{ccc}a_1+3r=-4\Big|\cdot(-1)\\a_1+7r=12\\\end{array}\right \Rightarrow \left\{\begin{array}{ccc}-a_1-3r=4\\a_1+7r=12\\\end{array}\right\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~----------\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~/~~~~4r=16 \Rightarrow r=\cfrac{16}{4} \Rightarrow \boxed{r=4}$

$2.~~2x^2-4x-1=0,~x_1+x_2+4x_1x_2=0\\\\ a=2,~b=-4,~c=-1\\\\ x_1+x_2=-\cfrac{b}{a}=-\cfrac{-4}{2} =2 \\\\ x_1x_2=\cfrac{c}{a} =\cfrac{-1}{2} =-\cfrac{1}{2} \\\\ x_1+x_2+4x_1x_2=2+4 \cdot \Bigg(-\cfrac{1}{2} \Bigg)=2-\cfrac{4}{2} =2-2=\boxed{0}$

$3.~~\sqrt[3]{3x-4}=2 \Rightarrow \Big(\sqrt[3]{3x-4} \Big)^3=2^3 \Rightarrow 3x-4=8 \Rightarrow 3x=8+4 \Rightarrow \\\\ \Rightarrow 3x=12 \Rightarrow x=\cfrac{12}{3}\Rightarrow \boxed{x=4}$

$4.~~A_5^3,~C_6^2,~4!\\\\A_5^3=\cfrac{5!}{(5-3)!} =\cfrac{5!}{2!} =\cfrac{\not1 \cdot \not2 \cdot 3 \cdot 4 \cdot 5}{\not1 \cdot \not2} =3 \cdot 4 \cdot 5=60 ~\vdots ~4\\\\C_6^2=\cfrac{6!}{2!(6-2)!} =\cfrac{6!}{2! \cdot 4!} =\cfrac{\not1 \cdot \not2\cdot \not3 \cdot \not4 \cdot 5 \cdot 6}{\not1 \cdot \not2 \cdot 1 \cdot 2 \cdot \not3 \cdot \not4} =\cfrac{30}{2} =15\not \vdots~4\\\\4!=1 \cdot 2 \cdot 3 \cdot 4=24~\vdots ~4$

$Cazuri~favorabile:~A_5^3,~4! \rightarrow 2~numere.\\\\Cazuri~posibile:~A_5^3,~C_6^2,~4!\rightarrow 3~numere.\\\\ P=\cfrac{nr.~cazuri~favorabile}{nr.~cazuri~posibile} \Rightarrow \boxed{P=\cfrac{2}{3} }$

$5.~~A(4,3),~B(6,7),~C(7,4)\\\\ Notam~cu~M(x_M,y_M)~mijlocul~segmentului~AB.\\\\ x_M=\cfrac{x_A+x_B}{2} =\cfrac{4+6}{2} =\cfrac{10}{2} =5\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~M(5,5)\\y_M=\cfrac{y_A+y_B}{2} =\cfrac{3+7}{2} =\cfrac{10}{y2} =5$

$CM=\sqrt{(x_M-x_C)^2+(y_M-y_C)^2} \Rightarrow CM=\sqrt{(5-7)^2+(5-4)^2}\Rightarrow \\\\ \Rightarrow CM=\sqrt{(-2)^2+1^2} \Rightarrow CM=\sqrt{4+1}\Rightarrow \boxed{CM=\sqrt{5} }$

$6.~~sinx=\cfrac{4}{5} , x\in\Bigg(\cfrac{\pi}{2} ,\pi\Bigg),~tgx=?\\\\ tgx=\cfrac{sinx}{cosx} \\\\ sin^2x+cos^2x=1 \Rightarrow cosx=\pm\sqrt{1-sin^2x} \\\\ x\in\Bigg(\cfrac{\pi}{2},\pi\Bigg) \Rightarrow cosx= -\sqrt{1-sin^2x} \Rightarrow cosx=-\sqrt{1-\Bigg(\cfrac{4}{5} \Bigg)^2} \Rightarrow$

$\Rightarrow cosx=-\sqrt{1-\cfrac{16}{25} } \Rightarrow cosx=-\sqrt{\cfrac{25-16}{25} } \Rightarrow cosx=-\sqrt{\cfrac{9}{25} }\Rightarrow \\\\ \Rightarrow cosx=-\cfrac{\sqrt{9} }{\sqrt{25} } \Rightarrow cosx=-\cfrac{3}{5}$

$tgx=\cfrac{sinx}{cosx} \Rightarrow tgx=\cfrac{\cfrac{4}{5} }{-\cfrac{3}{5} } \Rightarrow tgx=\cfrac{4}{5} \cdot \Bigg(-\cfrac{5}{3} \Bigg)\Rightarrow tgx=-\cfrac{20}{15} \Rightarrow \boxed{tgx=-\cfrac{4}{3} }$