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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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Answer:

<em>2.5</em>

Step-by-step explanation:

A direct variation equation has the form y = kx, where k is the constant of proportionality.

In this case, the constant of proportionality is 2.5.

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7 0
3 years ago
Element X decays radioactively with a half life of 12 minutes. If there are 200 grams of Element X, how long, to the nearest ten
Semenov [28]

Answer:

It would take 24 minutes for the element to decay to 50 grams

Step-by-step explanation:

The equation for the amount of the element present, after t minutes, is:

Q(t) = Q(0)e^{-rt}

In which Q(X) decays radioactively with a half life of 12 minutes.(0) is the initial amount and r is the rate it decreases.

Half life of 12 minutes

This means that Q(12) = 0.5Q(0)

So

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-12r}

e^{-12r} = 0.5

\ln{e^{-12r}} = \ln{0.5}

-12r = \ln{0.5}

12r = -\ln{0.5}

r = -\frac{\ln{0.5}}{12}

r = 0.05776

If there are 200 grams of Element X, how long, to the nearest tenth of a minute, would it take the element to decay to 50 grams?

This is t when Q(t) = 50. Q(0) = 200.

Q(t) = Q(0)e^{-rt}

50 = 200e^{-0.05776t}

e^{-0.05776t} = 0.25

\ln{e^{-0.05776t}} = \ln{0.25}

-0.05776t = \ln{0.25}

0.05776t = -\ln{0.25}

t = -\frac{\ln{0.25}}{0.05776}

t = 24

It would take 24 minutes for the element to decay to 50 grams

6 0
3 years ago
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