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Zolol [24]
3 years ago
14

Simplify

ass="latex-formula">
_____________
\sqrt[3]{4}
Mathematics
1 answer:
professor190 [17]3 years ago
3 0

Taking the square root of a number means finding a number that when you square it, it equals the number in the square root.

√4 = 2 because 2² = 4

³√4 is already simplified.

Best of Luck!

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Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?
Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

------------------------

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ =  sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

------------------------

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0
3 years ago
Please help . I’ll mark you as brainliest if correct
Oxana [17]

Answer:

30m

Step-by-step explanation:

51-21=30 that is the length of the long piece

8 0
3 years ago
Can someone please help me with #12?
Genrish500 [490]

4x+12=90

4x=90-12

4x=8

x=8/4

x=2

5 0
3 years ago
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