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Preeti Dabral 2 years, 3 months ago
According to the question,
Radius of the cones= r
Height of the upper cone= h
Height of the lower cone= H
Volume of the upper cone =13Ï€r2h
Volume of the lower cone =13Ï€r2H
Total volume of both the cones =13Ï€r2h+13Ï€r2H
=13Ï€r2(h+H)
The quantity of water displaced by the cones = The total volume of both the cones
therefore,
The quantity of water displaced will be 13Ï€r2(h+H) units 3
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Preeti Dabral 2 years, 3 months ago
Here,wehave⇒(x−3)(x−4)=34/332⇒x2−7x+12−34/332=0⇒x2−7x+13034/332=0⇒x2−7x+98/33×133/33=0⇒x2−231/33x+98/33×133/33=0
Byusingfactorizationmethod,weget⇒x2−(98/33+133/33)x+98/33×133/33=0⇒x2−98/33x−133/33x+98/33×133/33=0⇒(x2−98/33x)−(133/33x−98/33×133/33)=0⇒x(x−98/33)−133/33(x−98/33)=0⇒(x−98/33)(x−133/33)=0⇒x−98/33=0 or x−133/33=⇒x=98/33,133/33 Hence, x=x=98/33,133/33
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Preeti Dabral 2 years, 3 months ago
Let α and β are the zeroes of the polynomial f(x)=ax2+bx+c.
∴(α+β)=−baandαβ=caAccordingtothequestion,1αand1βarethezeroesoftherequiredquadraticpolynomial
Sum of zeroes of required polynomial
S′=1α+1β=α+βαβ=−bc[Fromequation(i)and(ii)]andproductofzeroesofrequiredpolynomial=1α×1β.P′=1αβ=ac[Fromequation(ii)]
Equation of the required polynomial.
=k(x2−S′x+p′)where$k$isanynon−zeroconstant=k(x2−(−bc)x+ac)[Fromequation(iii)and(iv)]=k(x2+bcx+ac)
Posted by Shanmuga Kani 2 years, 3 months ago
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Preeti Dabral 2 years, 3 months ago
The given equations are
71x+37y=253.........(i)
37x+71y=287.........(ii)
Adding (i) and (ii),we get
108x+108y=540
108(x+y)=540
∴x+y=540108=5......(iii)
Subtracting (ii) from (i),
34x−34y=253−287=−34
34(x−y)=−34
∴x−y=−3434=−1....(iv)
Adding (iii) and (iv)
2x = 5 - 1 = 4
⇒x=2
Subtracting (iv) from (iii)
2y = 5 + 1 = 6
⇒y=3
∴ Solution is x = 2, y = 3
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Kartik Upreti 2 years, 3 months ago
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