Question

if H is inversely proportional to (2p-3)^3 and H=-5 when p=1 find (i) the value of H when p=2.5 (ii)the value of p when H= 5/27

Answer 1

Answer:

( i )  if  p = 2.5 , ∴ H = 0.625   ( ii ) if  $H = \frac{5}{27}$    ,    ∴ p = 0

Step-by-step explanation:

Given,

H is inversely proportional to $(2p-3)^{3}$

i.e,  H = $\frac{K}{(2p-3)^{3}}$

If  H=-5 and p=1

then,

    $K = H(2p-3)^{3}$

⇒ K = $5 (2*1-3)^3$

⇒ K = $5(-1)^3$

∴  K = - 5

Now

( i )  if  p = 2.5

Then,     $H = \frac{K}{(2p-3)^{3}}$

        ⇒ $H = \frac{-5}{(2*2.5-3)^{3}}$

       ⇒ $H = \frac{-5}{(2)^{3}}$

       ⇒ $H = \frac{-5}{8}$

        ∴ H = 0.625

( ii ) if  $H = \frac{5}{27}$

Then, $H = \frac{K}{(2p-3)^{3}}$

        ${(2p-3)^{3}} = \frac{K}{H}$

        ${(2p-3)^{3}} = \frac{-5}{\frac{5}{27} }$

        ${(2p-3)^{3}} = - 27$

        ${(2p-3)^{3}} = - 3^{3}$

Here power are same, so equating the base

       (2p - 3) = - 3

       ∴ p = 0