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Ivenika [448]
3 years ago
11

Use the Rational Root Theorem to list all possible rational roots for the equation.

Mathematics
2 answers:
Anna71 [15]3 years ago
8 0
Q^n + k^(n-1).... + P = 0
Like: 3x^3+9x-6=0

All rational roots will be rational factors of P/Q such that:

Q = 3 Factors: 1,3
P = -6 Factors: [+/-] 1,2,3,6

Possible Rational Roots: [1/1,1/3,2/1,2/3,3/1,3/3,6/1,6/3] = [+/-] 1,1/3,2,2/3,3,6

Now you just test them in the equation itself and where the input makes the function equal 0, you have a root.

For this polynomial, no roots are rational, so when you test it you'll find that it must only contain irrational roots and may only be solved by other means.
Mariana [72]3 years ago
6 0

\bf \stackrel{q}{3}x^3+9x-\stackrel{p}{6}~\hspace{5em}\stackrel{\textit{factors of \boxed{p}}}{3,2,1,6}\qquad \stackrel{\textit{factors of \boxed{q}}}{3,1}\qquad \qquad \pm\cfrac{p}{q} \\\\\\ \textit{therefore, using the \underline{rational root test}}


\bf \begin{cases} \pm \cfrac{3}{3}\implies &\pm 1\\\\ \pm \cfrac{3}{1}\implies &\pm 3\\\\ \end{cases}\quad \begin{cases} \pm \cfrac{2}{3}\\\\ \pm \cfrac{2}{1}\implies \pm 2 \end{cases}\quad \begin{cases} \pm \cfrac{1}{3}\\\\ \pm \cfrac{1}{1}\implies \pm 1 \end{cases}\quad \begin{cases} \pm \cfrac{6}{3}\implies &\pm 2\\\\ \pm \cfrac{6}{1}\implies &\pm 6 \end{cases}

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m/s --- km/h = \frac{1000}{3600}

km/h= \frac{5}{18}

6m/s= \frac{6*5}{18}

6m/s= 1.66 km/h

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How do you know when a question is a division problem
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3 years ago
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A cylindrical tank with radius 3 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increas
Fittoniya [83]

Height of the water increasing is at rate of  #(dh)/(dt)=3/(25 pi)m/(min)#

<h3>How to solve?</h3>

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:

#V=pi r^2 h#

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:

#V=pi (5m)^2 h#

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:

#(dV)/(dt)=(25 m^2) pi (dh)/(dt)#

In the problem, we are given #3(m^3)/min# which is #(dV)/(dt)#.

So we need to substitute this in:

#(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)#

Hence,  Height of the water increasing is at rate of  #(dh)/(dt)=3/(25 pi)m/(min)#

<h3>Formula used: </h3>

#V=pi r^2 h#

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2 years ago
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Calvin had 30 minutes in time-out. For the first 23 1/3 minutes, Calvin counted spots on the ceiling. For the rest of the time,
scoundrel [369]

Answer: \frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

Step-by-step explanation:

For this exercise you can convert the mixed number to an improper fraction:

1. Multiply the whole number part by the denominator of the fraction.

2. Add the product obtained and the numerator of the fraction (This will be the new numerator).

3. The denominator does not change.

Then:

23\frac{1}{3}= \frac{(23*3)+1}{3}= \frac{70}{3}\ minutes

You know that he had 30 minutes in time-out, he counted spots on the ceiling for \frac{70}{3} minutes and the rest of the time he made faces at his stuffed tiger.

Then, in order to calculate the time Calvin spent making faces at his stuffed tiger, you need to subract 30 minutes and \frac{70}{3} minutes:

30\ min-\frac{70}{3}=(\frac{3(30)-70}{3})=\frac{20}{3}\ minutes or 6 \frac{2}{3}\ minutes

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