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Arisa [49]
3 years ago
11

Show that (4p^2-3q)^2+48p^2q =(4p^2+3q^2)​

Mathematics
1 answer:
Brut [27]3 years ago
5 0

Given:

The statement is

(4p^2-3q)^2+48p^2q=(4p^2+3q^2)

To prove:

The given statement.

Solution:

We have,

(4p^2-3q)^2+48p^2q=(4p^2+3q^2)

Now,

LHS=(4p^2-3q)^2+48p^2q

Using (a-b)^2=a^2+b^2-2ab, we get

LHS=(4p^2)^2+(3q)^2-2(4p^2)(3q)+48p^2q

LHS=(4p^2)^2+(3q)^2-24p^2q+48p^2q

LHS=(4p^2)^2+(3q)^2+24p^2q

It can be rewritten as

LHS=(4p^2)^2+(3q)^2+2(4p^2)(3q)

Using (a+b)^2=a^2+b^2+2ab, we get

LHS=(4p^2+3q^2)

LHS=RHS

Hence proved.

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The answer to your question is below

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A game decreased in price by 1/5. After the reduction it was priced at £48. What was the original price of the game?
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Answer:

1/5 is equal to 20% so you multiply 48 by .20 and you get 9.6 and your take 9.6 and add it to 48. your answer will be 57.6

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