3.12=
-12 +3.12
giúp mik bài 3.12
\(t=16\) phút \(5\) giây \(=965s\)
a. Có:
\(R_{12}=R_1+R_2=2+4=6\Omega\)
\(R_N=\dfrac{R_{12}R_3}{R_{12}+R_3}=\dfrac{6.12}{6+12}=4\Omega\)
\(I=\dfrac{E}{R_N+r}=\dfrac{13,5}{4+2}=2,25A\)
\(I_2=\dfrac{R_3}{R_3+R_{12}}.I=\dfrac{12}{12+6}.2,25=1,5A\)
Khối lượng bạc:
\(m=\dfrac{AIt}{nF}=\dfrac{108.1,5.965}{96500.1}=1,62g\)
b. Có:
\(I_{dm}=I_2=1,5A\)
\(U_{dm}=I_{dm}.R=1,5.4=6V\)
\(P_{dm}=I_{dm}.U_{dm}=1,5.6=9W\)
Giúp tớ bài 3.12
4.15-34.4+4.4/3.6-3.12
Tìm x : x.3=3.12=36
x=12
tick mình nhé rồi mình tick lại cho!
3 mũ 3.18-3 mũ 3.12
33 x 18 - 33 x 12
= 27 x 18 - 27 x 12
= 486 - 27 x 12
= 486 - 324
= 162
nha bn
S=1.6+2.9+3.12+4.15+.....+98.297+99.300
Theo đầu bài ta có:
\(S=1\cdot6+2\cdot9+3\cdot12+4\cdot15+...+98\cdot297+99\cdot300\)
\(=1\cdot\left(2\cdot3\right)+2\cdot\left(3\cdot3\right)+...+98\cdot\left(99\cdot3\right)+99\cdot\left(100\cdot3\right)\)
\(=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+98\cdot99\cdot\left(100-97\right)+99\cdot100\cdot\left(101-98\right)\)
\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+98\cdot99\cdot100-97\cdot98\cdot99+99\cdot100\cdot101-98\cdot99\cdot100\)
\(=99\cdot100\cdot101\)
\(=999900\)
\(3.8.5^2+2.4^3.12+\left(2^3+3\right).6.4\)
\(=3\cdot2^3\cdot5^2+2\cdot2^6\cdot2^2\cdot3+2^3\cdot2\cdot3\cdot2^2+3\cdot2\cdot3\cdot2^2\\ =2^3\cdot3\cdot5^2+2^9\cdot3+2^6\cdot3+2^3\cdot3^2\\ =2^3\cdot3\left(5^2+2^6+2^3+3\right)=24\left(25+64+8+3\right)\\ =24\cdot100=2400\)
4×50.08×3.12-50.08×1.48-50.08
4 x 50,08 x 3,12 - 50,08 x 1,48 - 50,08
= 50,08 x ( 4 x 3,12 ) - 50,08 x ( 1,48 - 1 )
= 50,08 x 12,48 - 50,08 x 0,48
= 50,08 x ( 12,48 - 0,48 )
= 50,08 x 12
= 600,69
Vậy 4 x 50,08 x 3,12 - 50,08 x 1,48 - 50,08 = 600,69