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skelet666 [1.2K]
3 years ago
14

According to the rational root theorem, which of the following are possible

Mathematics
1 answer:
White raven [17]3 years ago
7 0

Given:

The polynomial function is

F(x)=4x^3-6x^2+9x+10

To find:

The possible roots of the given polynomial using rational root theorem.

Solution:

According to the rational root theorem, all the rational roots and in the form of \dfrac{p}{q}, where, p is a factor of constant and q is the factor of leading coefficient.

We have,

F(x)=4x^3-6x^2+9x+10

Here, the constant term is 10 and the leading coefficient is 4.

Factors of constant term 10 are ±1, ±2, ±5, ±10.

Factors of leading term 4 are ±1, ±2, ±4.

Using rational root theorem, the possible rational roots are

x=\pm 1+,\pm 2, \pm 5, \pm 10,\pm \dfrac{1}{2}, \pm \dfrac{5}{2}, \dfrac{1}{4}, \pm \dfrac{5}{4}

Therefore, the correct options are A, C, D, F.

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three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

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3 years ago
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Answer:

3.  9x+5+x-7=25

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Step-by-step explanation:

3.  9x+5+x-7=25

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Step-by-step explanation:

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