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Ekta Tiwari 6 years, 4 months ago
Posted by Vijay Singh Khetwal 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let {tex}\alpha{/tex} and {tex}6\alpha{/tex} be the roots of equation.
We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8
Sum of zeroes{tex} = -\frac ba = -\frac{-14}{p}{/tex}
{tex}\alpha +6\alpha=\frac{14}{p}{/tex}
{tex}7\alpha = \frac{14}{p}{/tex}
{tex}\alpha = \frac2p{/tex}............(i)
Also, Product of the zeroes {tex} = \frac 8p=\frac ca{/tex}
{tex}\alpha \times 6\alpha = \frac 8p{/tex}
{tex}6\alpha^2=\frac 8p{/tex}
From (i)
{tex}6(\frac{2}{p})^2=\frac 8p{/tex}
{tex}6\times \frac {4}{p^2}=\frac 8p{/tex}
{tex}\frac{6}{p^2}=\frac2p{/tex}
{tex}\frac 62=\frac{p^2}{p}{/tex}
{tex}Hence, \ p=3{/tex}
Posted by Roshan Kumrawat 6 years, 4 months ago
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Posted by Varsha Borate 6 years, 4 months ago
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Abdul Hamid Hamid 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Check revision notes to check formulae : https://mycbseguide.com/cbse-revision-notes.html
Posted by Mir Muzaffar 6 years, 4 months ago
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Posted by Geetansh Dadani 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
2x +3y =11 ............... (1)
2x−4y =−24 ................ (2)
Using equation (2), we can say that
2x =−24 + 4y
⇒ x =−12 + 2y
Putting this in equation (1), we get
2(−12 + 2y )+3y =11 ⇒ −24 + 4y + 3y =11
⇒ 7y =35 ⇒ y =5
Putting value of y in equation (1), we get
2x +3(5)=11 ⇒ 2x +15=11
⇒ 2x =11 - 15=−4 ⇒ x =−2
Therefore, x =−and y =5
Putting values of x and y in y =mx +3, we get
5=m (−2) + 3 ⇒ 5 =−2m +3
⇒ −2m =2 ⇒ m =−1
Posted by Santu Shingh Santu Shingh 6 years, 4 months ago
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Posted by Airaf Khan 6 years, 4 months ago
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Harshita Talib 6 years, 4 months ago
Posted by Shivang Pratap Singh 6 years, 4 months ago
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Posted by Dipali Baruah 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given the base radius of the cone and the hemisphere are equal.
Diameter of hemisphere = 7 cm
{tex}\Rightarrow{/tex}Radius of hemisphere = 3.5 cm
Radius of the cone = Radius of the hemisphere = 3.5 cm
Let H be the total height of the top.
H = h + r
H = h + 3.5 ...(where h is the height of the cone.) .....(i)
Now,
Volume of toy = Volume of cone + Volume of hemisphere
{tex}\Rightarrow 231 = \left( \frac { 1 } { 3 } \pi r ^ { 2 } h \right) + \left( \frac { 2 } { 3 } \pi r ^ { 3 } \right){/tex}
{tex}\Rightarrow 231 = \frac { 1 } { 3 } \pi r ^ { 2 } ( h + 2 r ){/tex}
{tex}\Rightarrow 231 = \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 3.5 \times 3.5 ( h + 2 \times 3.5 ){/tex}
{tex}\Rightarrow 231 = \frac { 269.5 } { 21 } ( h + 7 ){/tex}
{tex}\Rightarrow h + 7 = \frac { 4851 } { 269.5 }{/tex}
{tex}\Rightarrow{/tex} h + 7 = 18
{tex}\Rightarrow{/tex} h = 11 cm
{tex}\Rightarrow{/tex} Height of the toy = h + r
= 11 + 3.5 ...(From (i))
= 14.5 cm
Posted by Ramprakash Jaiswal 6 years, 4 months ago
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Jatin Virwani 6 years, 4 months ago
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Sia ? 6 years, 3 months ago
Given polynomial is p(x) = 2x3 - x2- 5x - 2
and -1 and 2 are zeroes of polynomial.
{tex}\therefore{/tex} {x - (-1)} (x - 2)= ( x + 1) (x - 2) = x2 - 2x + x - 2 = x2- x - 2 is a factor of p(x)
For other zeroes, 2x + 1 = 0
{tex}\Rightarrow x = \frac { - 1 } { 2 }{/tex}
{tex}\therefore{/tex} Other zero = {tex}\frac { - 1 } { 2 }{/tex}
Posted by Shubhi Gupta 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
According to the question,we have to find the value of a such that the product of the zeros of the polynomial (ax2 - 6x - 6) is 4.
Let {tex} \alpha{/tex} and {tex}\beta{/tex} be the zeros of the polynomial (ax2 - 6x - 6)
Then, {tex}\alpha{/tex}{tex} \beta{/tex} = {tex} \frac { \text { constant term } } { \text { coefficient of } x ^ { 2 } } = \frac { - 6 } { a }{/tex}
But, {tex} \alpha{/tex}{tex} \beta{/tex} = 4 (given).
{tex} \therefore \quad \frac { - 6 } { a } = 4 \Rightarrow 4 a = - 6 \Rightarrow a = \frac { - 6 } { 4 } = \frac { - 3 } { 2 }{/tex}
Hence, a = {tex} \frac { - 3 } { 2 }{/tex}
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Aditi Mishra 6 years, 4 months ago
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