Question

,if x=10 find y
say answer ​

Answer 1

Answer:

5)x=2/5 and x= -2/3

6)y=4

Step-by-step explanation:

5)Given- P(x)= 15x²+4x-4

To find the zeroes of polynomial.

Sol: P(x)= 15x²+4x-4

A=+4

M=15×(-4)

=15x²+10x-6x-4

=5x(3x+2)-2(3x+2)

=(5x-2)(3x+2)=0

=x=2/5 and x=-2/3

6)Given- x is proportional to 1/√y when x=40, y=16 and if x=10

Then find the value of y

Sol: 40:16::10:y

40y=160

y=160/40

y= 4

Hope this Help you....

Answer 2

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \: \large \orange{ \bold{ \underline{ \underline{ \: step \: by \: step \: explaination : }}}}$

$\: \bf{(3) \: factorise : 15 {x}^{2} + 4x - 4 = 0}$

$\: \qquad \implies \displaystyle \sf{15 {x}^{2} + 4x - 4 = 0}$

$\: \: \\ \qquad \implies \displaystyle \sf{ 15 {x}^{2} + 10x - 6x - 4 = 0}$

$\: \: \\ \qquad \implies \displaystyle \sf \: 5x(3x + 2) - 2(3x + 2) = 0$

$\: \: \qquad \implies \displaystyle \sf (3x + 2)(5x - 2) = 0$

$\: \: \: \qquad \implies \displaystyle \sf \: x = - \frac{2}{3} \: \: \: and \: x = \frac{2}{5}$

$\: \: \: \qquad \bigstar \boxed{ \underline{ \rm{roots \: are \: x = \frac{ - 2}{3} \: \: and \: x = \frac{2}{5} }}}$

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$\: \bf{(2) x \propto} \: \frac{1}{ \sqrt{y} } \: \: \: when \: \: x =40 \: and \: \: y = 16$

$\: \bf{if \: x = 10 \: then \: y = to \: find}$

$\: \implies \displaystyle \sf{x \propto \frac{1}{ \sqrt{y} } }$

$\: \: \implies \displaystyle \sf \: 40 \propto \: \frac{1}{ \sqrt{16} }$

$\: \implies \displaystyle \sf \: 40 \propto \frac{1}{4}$

$\: \: \implies \displaystyle \sf \: 10 \propto \: \frac{1}{ \sqrt{y} }$

$\: \: \\ \implies \displaystyle \sf \therefore \: x = \frac{k}{ \sqrt{y} } \: where \: k \: is \: constant \: of \: variation$

$\: \: \implies \displaystyle \sf \: x \times \sqrt{y} = k$

$\: \: \implies \displaystyle \sf \: when \: x = 40 \: and \: y = 16$

$\: \implies \displaystyle \sf \: \therefore \: k = 40 \times \sqrt{16}$

$\: \implies \displaystyle \sf \: 40 \times 4 = 160$

$\: \: \\ \implies \displaystyle \sf \: so \: the \: equation \: of \: variation \: is$

$\: \: \implies \displaystyle \sf \: x \sqrt{y} = 160$

$\: \implies \displaystyle \sf \: when \: x = 10$

$\: \: \implies \displaystyle \sf \: 10 \sqrt{y} = 160$

$\: \: \implies \displaystyle \sf \: \sqrt{y} = \frac{160}{10} = \cancel \frac{160}{10} = \frac{16}{1}$

$\: \: \implies \displaystyle \sf \: \sqrt{y} = 16$

$\: \: \implies \displaystyle \sf \: y = {(16)}^{2}$

$\: \: \implies \displaystyle \sf \: y = 256$

$\: \: \bigstar \boxed{ \boxed{ \underline {\bf{the \: value \: of \: y \: is \: 256}}}}$

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