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2 years ago

Find the length of side xx in simplest radical form with a rational denominator.

Mathematics

1 answer:

anygoal [31]2 years ago

8 0

Answer:

$x = \frac{3}{ \sqrt{2} }$

Step-by-step explanation:

${x}^{2} + {x}^{2} = {3}^{2} \\ 2 {x}^{2} = 9 \\ {x}^{2} = \frac{9}{2} \\ x = \frac{3}{ \sqrt{2} } \\ x = 2.121$

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All of the picture please help

Sonbull [250]

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B. Equation: 4 cups multiplied by 4(four quarter cups per cup) = 16 quarter cups

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5 0

3 years ago

Is 1/4 equivalent to 2/7?

Mazyrski [523]

No they are not equivalent

6 0

3 years ago

Read 2 more answers

Find the following measure for this figure.

miv72 [106K]

l=11
r=5

Now

LSA:-

πrl
π(11)(5)
55πunits ²

4 0

2 years ago

Factor 140c + 28 - 14a to identify the equivalent expressions.

Feliz [49]

Answer:

The answer to your question are letters A, C and D.

Step-by-step explanation:

To factor this polynomial we need to look for the common factor of the three terms.

To find it, get the greatest common factor

140 28 14 2

70 14 7 2

35 7 7 5

7 7 7 7

1 1 1

Greatest common factor = 2 x 7 = 14

Factor the polynomial

140c + 28 - 14a = 14 ( 10c + 2 - a)

Factor only 7 7(20c + 4 - 2a)

Factor only 2 2 (70c + 14 - 7a)

8 0

2 years ago

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

2 years ago