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Mars2501 [29]
3 years ago
9

How do I solve 2x^3-32x=0

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0
We are asked to solve the equation 2x³ - 32x = 0. First off, we can see that the variable x appears in both terms on the left-hand side of the equation. Therefore, we can factor it out. We can also factor out a common factor of 2 from both terms. 
2x³ - 32x = 0
2x(x² - 16) = 0
We can use the difference of squares pattern to further simplify the equation.
2x(x + 4)(x - 4) = 0
Now, using the Zero Product Property, set each term to zero.
2x = 0 
x = 0
x + 4 = 0
x = -4
x - 4 = 0
x = 4
Therefore, the solutions to the equation 2x³ - 32x are 0, -4, and 4. Hope this helped and have a great day!
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A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y
siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32


Given that 100 beetles are released into a field.

So that means initial number of beetles P=100


Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

A=P(1+r)^n where A is number of beetles at any year n.

Plug the given values into above formula we get:

A=100(1+0.32)^n

A=100(1.32)^n


now plug n=10 years

A=100(1.32)^{10}=100(16.0597696605)=1605.97696605

Hence answer is approx 1606 beetles will be there after 20 years.


Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

A=100(1.32)^{20}=100(257.916201549)=25791.6201549

Hence answer is approx 25791 beetles will be there after 20 years.



Now we have to find time for 100000 beetles so plug A=100000

A=100(1.32)^n

100000=100(1.32)^n

1000=(1.32)^n

log(1000)=n*log(1.32)

\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n

33.174666862=n

Hence answer is approx 33 years.

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