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algol [13]
3 years ago
5

Write the difference in factored form. (24x4 - 15x2 + 6x) - (10x4 + 5x2 - 4x)​

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

The factored difference is: 2x(7x^3 - 10x + 5)

Step-by-step explanation:

Difference in non factored form:

To find the difference in non-factored form, we subtract the common terms. So

24x^4 - 15x^2 + 6x - (10x^4 + 5x^2 - 4x) = 24x^4 - 10x^4 - 15x^2 - 5x^2 + 6x + 4x = 14x^4 - 20x^2 + 10x

Greatest common factor:

Of the exponents: Between x^4, x^2 and x, it is the one with the lowest exponent, so x.

Between 14, 20 and 10:

14 - 20 - 10|2

7 - 10 - 5

So 2

The gcf is 2x, which means that the factored expression is:

2x(\frac{14x^4}{2x} - \frac{20x^2}{2x} + \frac{10x}{2x}) = 2x(7x^3 - 10x + 5)

The factored difference is: 2x(7x^3 - 10x + 5)

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3 years ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
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Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
3 years ago
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