Question

AJUTORR !! DAU CORONITA !!
1) O minge de tenis are diametrul de 7,2 cm.
a) Calculati aria mingii
b) Aflati volumul mingii
c) Cate mingi de tenis incap intr-o cutie in forma de paralelipiped dreptunghic cu dimensiunile : L= 72cm; l=36cm si h=36cm ?
2) Amplificand fractia 3supra5 cu 5, obtinem fractia... .
3) Inversul numarului -5supra8 este numarul... .
4) Scriind ca interval multimea A={x∈R| 2-x>1}, obtinem A=... .
5) Aria dreptunghiului, avand lungimea de 10cm si latimea reprezetand 2supra5 din lungimea, este egala cu ... cm².
6) Verificati egalitatea a=2b, daca a=√72+√18-√2, iar b=8supra√2.
7) Se considera fractia f:R--->R, f(x)=-x-1.
a) Determinati m∈R, stiind ca punctul P(m,1) ∈ Gf.
8) Verficati egalitatea (x+3)²-4x-12=(x+3)(x-1), oricare ar fi numarul real x. 
AJUTATI-MA DAU CORONITA LA CEL MAI BUN RASPUNS :o33

Answer 1

$1.~D=7,2cm=2R \Rightarrow R=3,6cm. \\a) A= 4 \pi R^{2} =4 \pi 12,96=51,84 \pi ~(cm^{2}) \\ b)V= \frac{4 \pi R^{3}}{3}= \frac{4 \pi46,656 }{3}= 62,608~(cm^{3}) \\ c)~Pe~lungime~incap~[72:7,2]=10~mingi. \\ Pe~latime~incap~[36:7,2]=5~mingi. \\ Pe~inaltime~incap~[36:7,2]=5~mingi. \\ In~cutie~incap~10*5*5=250~mingi.$

$2.~ \frac{3}{5}= \frac{15}{25}. \\ 3.~ \frac{5}{8}. \\ 4.~2-x\ \textgreater \ 1~\Leftrightarrow 1\ \textgreater \ x \Rightarrow x \in (- \infty;1). \\ A=(-\infty;1). \\ 5.~L=10~cm \\ l= \frac{2}{5}*L= \frac{2}{5}*10=4~(cm). \\ A=L*l=10*4=40~(cm^2).$

$6.~a= \sqrt{72} + \sqrt{18} - \sqrt{2} =6 \sqrt{2}+3 \sqrt{2}- \sqrt{2}=8 \sqrt{2}. \\ b= \frac{8}{ \sqrt{2} }= \frac{8 \sqrt{2} }{2}=4 \sqrt{2} . \\ 2b=8 \sqrt{2}. \\ Deci~a=2b.$

$7.~P(m,1) \in G_{f} \Leftrightarrow f(m)=1 \Leftrightarrow -m-1=1 \Rightarrow \boxed{m=-2}. \\ 8.~(x+3)^{2}-4x-12==(x+3) ^{2} -4(x+3)=(x+3)(x+3-4)= \\ =(x+3)(x-1).$