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Virty [35]
3 years ago
15

Factor the following expression completely please put A,B,C,D as the answer thanks:)

Mathematics
2 answers:
Studentka2010 [4]3 years ago
8 0
64x^6-x^4=x^4(64x^2-1)=x^4\left[(8x)^2-1^2\right]=x^4(8x-1)(8x+1)\\\\\\\\a^2-b^2=(a-b)(a+b)
Vanyuwa [196]3 years ago
3 0
1)x^4(8x-1)^2
=x^4(64x^2-16x+1)
=64x^6-16x^5+x^4

2)(8x^3-x^2)^2
=64x^6-16x^5+x^4

3)x^4(8x-1)(8x+1)
=x^4(64x^2-1)
=64x^6-x^4

4)x^4(64x^2-1)
=64x^6-x^4
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(cot theta + cosec theta)/(cosec theta (1 + (cot theta/ cosec theta)).
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The given expression \dfrac{\frac{cos \theta}{sin \theta}  + \frac{1}{sin\theta} }{\frac{1}{sin \theta} (1+cos \theta)} =1 is true

<h3>Proving trigonometry identity</h3>

Given the expression

\frac{cot \theta + cosec \theta}{cosec \theta(1+\frac{cot \theta}{cosec \theta} )} =1

Note that:

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Substituting the given parameters into the formula;

\frac{\frac{cos \theta}{sin \theta}  + \frac{1}{sin\theta} }{\frac{1}{sin \theta} (1+cos \theta)} =1

Find the LCM of both the numerator and denominator

= \dfrac{\frac{cos \theta + 1}{sin \theta} }{(\frac{1+cos \theta}{sin \theta} )}\\

Divide the result to have:

= \frac{1+cos \theta}{sin \theta} \times \frac{sin \theta}{1+cos \theta}\\ = \frac{1}{1}\\ = 1 (Proved)

Learn more on proofs of trigonometry identity here: brainly.com/question/7331447

4 0
2 years ago
Read 2 more answers
What is a common factor of both x^2-9 and x^2+x-6
Leni [432]

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7 0
3 years ago
student tickets for a football game cost $2 each. adult tickets cost $4 each. ticket sales at last week's game totaled $2800. wr
ELEN [110]

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3 years ago
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One third of a number m= (one third)(a number m)= (13)(a variable)  = (13)

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4 0
3 years ago
Please help! I don't get it! It is confusing me!
julia-pushkina [17]
Lets solve the equalities first.
NP>MN
Cancel out the N
P>M
Next
MP<MN
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3 years ago
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