Question prove that 9^3/2 -3 *5^0 -(1/81)^-1/2 = 15 Pls help, correct answer, I’ll mark as brainliest

Answer: [tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex] Step-by-step explanation: The given expression is : [tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex] We need to prove that LHS is equal to RHS. Taking LHS, [tex]=9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}[/tex] [tex]=(3)^{2\times\dfrac{3}{2}}-3-(\dfrac{1}{9})^{2\times \dfrac{-1}{2}}\\\\=27-3-9\\\\=15\\\\=RHS[/tex] Hence, LHS = RHS. Log in to Reply

Answer:[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]

Step-by-step explanation:The given expression is :

[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]

We need to prove that LHS is equal to RHS.

Taking LHS,

[tex]=9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}[/tex]

[tex]=(3)^{2\times\dfrac{3}{2}}-3-(\dfrac{1}{9})^{2\times \dfrac{-1}{2}}\\\\=27-3-9\\\\=15\\\\=RHS[/tex]

Hence, LHS = RHS.