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Solution :
11. sol.
Let x = 0.235 bar on 35
⇒ x = 0. 23 53 53 535........( i )
Multiply both sides by 10,
⇒ 10x = 2. 35 35 35 35...... ( ii )
Multiply both sides by 100,
⇒ 1000 x = 235. 35 35 35...... ( iii )
On subtracting ( ii ) from ( iii )
1000 x - 10 x = 235 - 2
⇒ 990 x = 233
⇒ x = 233/990
Therefore, 0.325 bar on 35 = 233/990
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12. sol.
STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit.
2: Mark O the mid point of AC
3: Draw a semicircle with O as centre & OA as radius
4: At B draw a perpendicular BD.
5: BD = √9.3 unit
6: Now, B as centre, BD as radius, draw an arc, meeting the numberline at E.
Now, with BD or BE = √9.3 as radius , with 0 ( origin) of the number line as centre, draw an arc on the the number line, intersecting at point ‘k’. And this point ‘k' lies between integers 3 & 4, & represents √9.3
JUSTIFICATION:
BD = √ {(10.3/2)² - (8.3/2)²}
=> BD = √{10.3²- 8.3²)/4 }
=> BD = √{(10.3+8.3)(10.3–8.3)/4}
=> BD = √{18.6*2/4}
=> BF = √{37.2/4}
=> BD = √9.3 = BE
plzz check the above figure 1 for the representation on a number line.
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13. sol.
First We take R.H.S & use the Formula [( a-b)²= a²+b²-2ab] & simplify it then R.H.S becomes equal to L.H.S
R.H.S
⇒ 1/2×(x + y + z) (x²+ y²-2xy +y²+ z²-2yz+x²+z²-2xz)
[( a-b)²= a²+b²-2ab]
⇒ 1/2×(x + y + z) (2x²+ 2y²+2z²-2xy -2yz-2xz)
⇒ 1/2×(x + y + z) 2(x² + y²+ z² – xy – yz – xz)
=(x + y + z) (x² + y²+ z² – xy – yz – xz)
= x³+y³+z³-3xyz
= L.H.S
We know that,
[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]
L.H.S = R.H.S
[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]
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14. sol.
8x³ + 27y³ +36x²y + 54xy²
=>(2x)³ + (3y)³ + 3 (2x)² (3y) + 3 (2x)(3y)²
we know ,
(a + b)³ = a³ + b³ + 3a²b + 3ab²
use this
=>(2x + 3y)³
hence ,
factorization of given expression is (2x + 3y) , (2x + 3y) and (2x + 3y)
15. sol.
Plzz check the above figure 2 for answer
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