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Oksanka [162]
3 years ago
11

Given that n^2+2pn+p^2+q^2=r^2 has real roots show that r^2=q^2​

Mathematics
1 answer:
Anon25 [30]3 years ago
4 0

It doesn't seem true, and here's a counterexample: observe that the first three terms form a perfect square. You can rewrite the equation as

(n+p)^2+q^2=r^2

This is basically the Pythagorean theorem applied to a triangle with sides n+p, q and r. For example, pick:

n=1, p=2, q=4, r^5

The expression becomes

1^2+2\cdot 2\cdot 1+2^2+4^2=5^2 \iff 1+4+4+16=25 \iff 25=25

Which is true, even if

r^2\neq q^2

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Answer:

Option A) is correct.

Step-by-step explanation:

The final cost of a sale item is determined by multiplying the price on the tag by 75%.

So. if the final cost is represented by $F and the price on the tag is $t, then the relation between  F and t will be  

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5 0
3 years ago
Read 2 more answers
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Lerok [7]

Answer:

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