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Ask QuestionPosted by Sufiyan Ansari 6 years, 8 months ago
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Posted by Archana Shelke 6 years, 8 months ago
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Sia ? 6 years, 8 months ago
Applying the relationship
{tex}\style{font-family:'Times New Roman'}{m\;=K_H\;p\;}{/tex}
In the first case, {tex}\style{font-family:'Times New Roman'}{6.56\;\times\;10^{-2}g\;=\;K_H\;\times\;1\;bar}{/tex}
{tex}\style{font-family:'Times New Roman'}{\;K_H\;=6.56\;\times\;10^{-2}g\;/bar}{/tex}
In second case, {tex}\style{font-family:'Times New Roman'}{\;5.00\;\times\;10^{-2}g\;=6.56\;\times\;10^{-2}g/bar\;\times\;p}{/tex}
{tex}P = \frac{{5.00 \times {{10}^{ - 2}}g}}{{6.56 \times {{10}^{ - 2}}g\,ba{r^{ - 1}}}}{/tex} = 0.7621 bar
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Yogita Ingle 6 years, 8 months ago
Process of Creation of Money:The process of money creation by the commercial banks starts as soon as people deposit money in their respective bank accounts. After receiving the deposits, as per the central bank guidelines, the commercial banks maintain a portion of total deposits in form of cash reserves. The remaining portion left after maintaining cash reserves of the total deposits is then lend by the commercial bank to the general public in form of credit, loans and advances. Now assuming that all transactions in the economy are routed through the commercial banks, then the money borrowed by the borrowers again comes back to the banks in form of deposits. The commercial banks again keep a portion of the deposits as reserves and lend the rest. The deposit of money by the people in the banks and the subsequent lending of loans by the commercial banks is a never-ending process. It is due to this continuous process that the commercial banks are able to create credit money a multiple time of the initial deposits.
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Sia ? 6 years, 8 months ago
we know that ,
equation of plane passing through {tex}(x_1 ,y_1, z_1){/tex} having Direction ratios a, b,c is a {tex}(x - x_1) + b (y- y_1) + c (z - z_1) = 0{/tex}
Equation of plane passing through the point A (1, 2, 1) is given as
{tex}a (x - 1 ) + b (y- 2) + c (z - 1 ) = 0{/tex} .....(i)
Now, DR's of line PQ where {tex}P(1, 4, 2)\ and\ Q(2, 3, 5)\ are\ ( 2 -1, 3 - 4, 5-2)=(1, -1, 3).{/tex}
Plane (i) is perpendicular to line PQ.
{tex}\therefore{/tex} Direction ratios of plane (i) are (1, -1, 3) [{tex}\because{/tex} Direction ratios normal to the plane are proportional]
i.e. {tex}a = 1, b = -1, c = 3{/tex}
On putting values of a, b and c in Eq. (i), we get the required equation of plane as
{tex}1 (x- 1 ) -1 (y - 2) + 3 (z- 1 ) = 0{/tex}
{tex}x - 1 - y + 2 + 3 z - 3 = 0{/tex}
{tex}x - y + 3z - 2 = 0{/tex} ......(ii)
Now, the given equation of line is
{tex}\frac { x + 3 } { 2 } = \frac { y - 5 } { - 1 } = \frac { z - 7 } { - 1 }{/tex}....(iii)
Direction ratios of this line are {tex}(2, -1, -1){/tex} and passing through the point {tex}(-3, 5, 7){/tex}.
Direction ratios of normal to the plane (ii) are {tex}(1,-1, 3){/tex}.
To check whether the line is perpendicular to the plane , we use the condition {tex}a_1 a_2 + b_1 b_2 + c_1 c_2 = 0{/tex}
{tex}2 (1 ) -1 (-1) -1 (3) = 2 + 1 -3 = 0{/tex}
So, line (iii) is perpendicular to plane (i).
Distance of the point {tex}(-3, 5, 7){/tex} from the plane (ii)
{tex}\Rightarrow d = \left| \frac { ( - 3 ) ( 1 ) + ( 5 ) ( - 1 ) + 7 ( 3 ) - 2 } { \sqrt { ( 1 ) ^ { 2 } + ( - 1 ) ^ { 2 } + ( 3 ) ^ { 2 } } } \right|{/tex}{tex}\left[ \begin{array} { c } { \because \text { distance of the point } \left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right) } \ { \text { to the plane } a x + b y + c z + d = 0 \text { is } } \\ { d = \frac { \left| a x _ { 1 } + b y _ { 1 } + a _ { 1 } + d \right| } { \sqrt { a ^ { 2 } + b ^ { 2 } + c ^ { 2 } } } } \end{array} \right]{/tex}
{tex}= \left| \frac { - 3 - 5 + 21 - 2 } { \sqrt { 1 + 1 + 9 } } \right|{/tex}
{tex}= \left| \frac { 11 } { \sqrt { 11 } } \right| = \left| \frac { ( \sqrt { 11 } ) ^ { 2 } } { \sqrt { 11 } } \right|{/tex}
{tex}= \sqrt { 11 }units{/tex}
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Posted by Aiswarya Mahesh 6 years, 8 months ago
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Sia ? 6 years, 8 months ago
from array import *
array_num = array('i', [1, 3, 5, 3, 7, 1, 9, 3])
print("Original array: "+str(array_num))
array_num.reverse()
print("Reverse the order of the items:")
print(str(array_num))

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Abhay Kr Pandey 6 years, 8 months ago
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