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Sia ? 6 years, 6 months ago
Let speed of ritu in still water = x km/h
Speed of stream = y km/h
Then downstream speed of ritu = x + y
Upstream speed of ritu = x - y
According to question
{tex}\frac{20}{x+y}=2{/tex}
{tex}\Rightarrow{/tex} x +y = 10 .... (1)
Also,
{tex}\frac{4}{x-y}=2{/tex}
{tex}\Rightarrow{/tex} x - y = 2 ..... (2)
Add (1) and (2)
2x = 12
{tex}\Rightarrow{/tex} x = 6
From (1), we get
6 + y = 10
{tex}\Rightarrow{/tex} y = 4
Therefore, Speed of ritu in still water = 6km/h
Speed of stream = 4km/h
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Sia ? 6 years, 6 months ago
Diameter of semicircular sheet is 28 cm. It is bent to from an open conical cup. The radius of sheet becomes the slant height of the cup. The circumference of the sheet becomes the circumference of the base of the cone.
{tex}\therefore \;l = {/tex} Slant height of conical cup = 14 cm.
Let r cm be the radius and h cm be the height of the conical cup circumference of conical cup of the semicircular sheet
{tex}\therefore \;2\pi r = \pi \times 14 \Rightarrow r = 7cm{/tex}
Now, {tex}{l^2} = {r^2} + {h^2} \Rightarrow h = \sqrt {{l^2} - {r^2}}{/tex}
{tex}= \sqrt {{{(14)}^2} - {{(7)}^2}} = \sqrt {196 - 49}{/tex}{tex}= \sqrt {147} = 12.12cm{/tex}
{tex}\therefore{/tex} Capacity of the cup
{tex} = \frac{1}{3}\pi {r^2}h = \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 7 \times 12.12{/tex}
= 622.16 cm3
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Sia ? 6 years, 6 months ago
Let E be the midpoint of AB.
{tex}\therefore \quad \frac { x + 1 } { 2 } = 2{/tex} or x = 3
and {tex}\frac { y + ( - 4 ) } { 2 } = - 1{/tex} or, y = 2
or, B(3, 2)
Let F be the mid-point of AC.Then,
{tex}0=\frac{x_1+1}{2}{/tex} or {tex}x_1=-1{/tex}
and {tex}\frac { y _ { 1 } + ( - 4 ) } { 2 }{/tex} = -1 or, y1 = 2
or, C= (-1, 2)
Now the co-ordinates are A(1, - 4), B(3,2), C (-1,2)
Area of triangle
{tex}= \frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}= \frac { 1 } { 2 } [ 1 ( 2 - 2 ) + 3 ( 2 + 4 ) - 1 ( - 4 - 2 ) ]{/tex}
{tex}= \frac { 1 } { 2 } [ 0 + 18 + 6 ]{/tex}
= 12 sq units.
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Sia ? 6 years, 6 months ago
{tex}\frac{1}{2a + b + 2x}{/tex} = {tex}\frac{1}{2a}{/tex} + {tex}\frac{1}{b}{/tex} + {tex}\frac{1}{2x}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{2a + b + 2x}{/tex} - {tex}\frac{1}{2x}{/tex} = {tex}\frac{1}{2a}{/tex} + {tex}\frac{1}{b}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex} = {tex}\frac{b + 2a}{2a \times b}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex} = {tex}\frac{b + 2a}{2ab}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex} = {tex}\frac{1}{2ab}{/tex}
{tex}\Rightarrow{/tex} {tex}4x^2 + 2bx + 4ax = -2ab{/tex}
{tex}\Rightarrow{/tex} {tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}
{tex}\Rightarrow{/tex} {tex}2x(2x + b) + 2a(2x + b) = 0{/tex}
{tex}\Rightarrow{/tex} (2x + b)(2x + 2a) = 0
{tex}\Rightarrow{/tex} x = -{tex}\frac{b}{2}{/tex} or x = -a
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