Question

1.Arata ca fractia 5 la puterea 21 supra 12 la puterea 14 este subunitara
2.Stiimd ca suprafata unui dreptunghi este de 162 metri patrati si latimea este 50% din lungimea sa, calculeaza perimetrul dreptunghiului.
3.Stiai ca un cartof contine 15% amidon.Cati cartofi au fost necesari pentru a obtine 18kg de amidon, stiind ca intr-un kg sunt aproximativ 6 cartofi.
CORONITA, VOT 5 STELE,UN MULTUMESC SI DAU 50 PUNCTE

Answer 1

5^21 comparat cu 12^14
(5³)^7 comparat cu (12²)^7
125^7 comparat cu 144^7
7=7
125<144
deci 125^7<144^7 deci numarator<numitor, fractie subunitara


fie l , latimea si 2l, lungimea
 atunci aria este
2l*l=162m²
2l²=162
l²=81
l=9m
2l=18m
 Perimetrul =2( l+2l) =6l=6*9=54 m


nr kg cartofi *0,15=18 kg amidon
nr kg cartofi =18:0,15=1800:15=120 kg cartofi

1 kg cartofi...6 cartofi
120 kg cartofi....x cartofi
d.p.
x=120*6=720 cartofi

Answer 2

1)

[tex]\it \dfrac{5^{21}}{12^{14}}=\dfrac{5^{3\cdot7}}{12^{2\cdot7}} =\dfrac{(5^3)^7} {(12^2)^7} =\dfrac{125^7}{144^7} = \left(\dfrac{125}{144}\right)^7 \ \textless \ \left(\dfrac{144}{144}\right)^7 =1 \Longrightarrow \\ \\ \\ \Longrightarrow \dfrac{5^{21}}{12^{14}} \ \textless \ 1[/tex]

2)

[tex]\it \mathcal{A} =L\cdot \ell \Rightarrow 162= L\cdot \ell \Rightarrow L\cdot \ell =162 \ \ \ \ \ (1) \\ \\ \\ 50\% = \dfrac{ \ \ 50^{(50}}{100} = \dfrac{1}{2} \Rightarrow \ell =\dfrac{L}{2} \Rightarrow L = 2\cdot\ell \ \ \ \ \ (2) \\ \\ \\ (1),\ (2) \Rightarrow 2\cdot\ell\cdot\ell=162|_{:2} \Rightarrow \ell\cdot\ell=81\Rightarrow \ell=9\ m \ \ \ \ (3)[/tex]

[tex]\it (2), (3) \Rightarrow L = 2\cdot9 = 18\ m \ \ \ \ \ (4) \\ \\ \\ \mathcal{P} = 2(L+\ell) = 2(18+9) = 2\cdot27 = 54\ m [/tex]

3)

[tex]\it Not\breve{a}m\ \ c\ -\ cantitatea\ de \ cartofi \\ \\ \\ 15\%\ din\ c = 18 \Rightarrow \dfrac{15}{100} \cdot\ c = 18 \Rightarrow c= 18\cdot \dfrac{100}{15} = 18\cdot\dfrac{20}{3} = 6\cdot20=120kg \\ \\ \\ 120\cdot 6 = 720\ cartofi\ sunt\ necesari.[/tex]