Question

va rog ajutatima plsssss

Answer 1

$1)Ca \ numerele \ sa \ fie \ in \ progresie \ geometrica, \ trebuie \ ca \ cel \ din \\ mijloc \ sa \ fie \ egal \ cu \ media \ geometrica \ a \ vecinilor \ sai: \\ 36=\sqrt{4*x}, x \geq 0 =\ \textgreater \ 36^2=4*x=\ \textgreater \ x=\frac{36^2}{4} = \frac{36^2}{2^2}=(\frac{36}{2})^2 =\ \textgreater \ \\ =\ \textgreater \ x = 18^2=\ \textgreater \ x=324.$
 $2)(f \circ f)(x)=f(f(x)); \\ Avem \ f(x)=x+a, \ deci f(f(x))=f(x+a)=(x+a)+a =\ \textgreater \ \\ f(f(x))=x+2a; \\ Dar \ noi \ trebuie \ sa \ aflam \ a\in \mathbb{R} \ pt \ care (f\circ f)(x)=x, x \in \mathbb{R}. \\ Inlocuim \ cu \ ce \ am \ aflat \ si \ ne \ rezulta: \\ \Rightarrow x+2a=x |-x=\ \textgreater \ 2a=0|:2=\ \textgreater \ a=0 \in \mathbb{R}.$
$3)Avem \ ecuatia: 3^{-x+2}=\sqrt{3}; Scriem \ pe \ \sqrt{3} \ ca \ putere \ de \ 3: \\ \sqrt{3}=3^{\frac{1}{2}} \ dupa \ formula\\ \sqrt[n]{a^m}=a^\frac{m}{n}. \\ \Rightarrow 3^{-x+2}=3^{\frac{1}{2}} =\ \textgreater \ -x+2=\frac{1}{2}|*2=\ \textgreater \ -2x+4=1=\ \textgreater \ \\ =\ \textgreater \ -2x=-3|*(-1)=\ \textgreater \ 2x=3=\ \textgreater \ x=\frac{3}{2}$
4) Luam pe rand submultimi neordonate de cate 1, 2, si 3 elemente.
Cel mult inseamna (mai mic sau egal)
$C_{4}^{1}=4 \ submultimi \ cu \ 1 \ element; \\ C_4^{2}=\frac{4*3}{2}=2*3=\ \textgreater \ C_4^{2}=6 \ submultimi \ cu \ 2 \ elemente; \\ C_4^{3}=C_{4}^{4-3}=C_{4}^{1}=4 \ submultimi \ cu \ 3 \ elemente; \\ In \ total: \ Avem \ 4+6+4=10+4=14 \ submultimi \ cu \ cel \ mult\\ 3 \ elemente.$
5)$\ Fie \ dreapta \ a \bot d \ si \ A(2,-3)\in a =\ \textgreater \ m_a*m_d=-1; \\ m_d=-\frac{a}{b} = - \frac{2}{1}=\ \textgreater \ m_d=-2; =\ \textgreater \ m_a=\frac{-1}{m_d}=\frac{-1}{-2} =\ \textgreater \ m_a=\frac{1}{2}; \\ Aplici \ ecuatia \ dreptei \ care \ trece \ printr-un \ punct \ si \ are \ panta \\ data: a:y-y_A=m_a(x-x_A) =\ \textgreater \ a:y+3=\frac{1}{2}(x+3)|*2 =\ \textgreater \ \\ =\ \textgreater \ a:2y+6=x+3=\ \textgreater \ a:x-2y-3=0.$
6) Aflam pe sin x din teorema fundamentala a trigonometriei:
$sin^2x+cos^2x=1=\ \textgreater \ sin^2x=1-cos^2x=1-(\frac{3\sqrt{5}}{7})^2 = 1 - \frac{45}{49} \\ Aducem \ la \ acelasi \ numitor: sin^2x=\frac{49-45}{49}=\ \textgreater \ sin^2x=\frac{4}{49} \\ sin \ x \ in \ cadranul \ I \ (0,\frac{\pi}{2}) \ este \ pozitiv\ deci: \\ sin \ x = \frac{2}{7}; \\ sin \ 2x=2 \ sin x \ cos x =\ \textgreater \ sin \ 2x = 2 * \frac{2}{7} * \frac{3\sqrt{5}}{7} =\ \textgreater \ sin \ 2x = \frac{12\sqrt{5}}{49}$