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Mathematics

1 answer:

SOVA2 [1]2 years ago

4 0

Answer:

Step-by-step explanation:

Let's plug in 105 for v. We get 105 = 30 $\sqrt[3]{t}$ .

3.5 = $\sqrt[3]{t}$

t = 43 seconds

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How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove

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The equation has one extraneous solution which is n ≈ 2.38450287.

Given that,

The equation;

$\dfrac{9}{n^2+1} =\dfrac{n+3}{4}$

We have to find,

How many extraneous solutions does the equation?

According to the question,

An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.

To solve the equation cross multiplication process is applied following all the steps given below.

$\rm \dfrac{9}{n^2+1} =\dfrac{n+3}{4}\\\\9 (4) = (n+3) (n^2+1)\\\\36 = n(n^2+1) + 3 (n^2+1)\\\\36 = n^3+ n + 3n^2+3\\\\n^3+ n + 3n^2+3 - 36=0\\\\n^3+ 3n^2+n -33=0\\$