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tiny-mole [99]
3 years ago
6

WILL MARK BRAINLIEST ONLY HAVE 10 MINUTES

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

A

Step-by-step explanation:

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Sales tax on an item is directly proportional to the cost of the item purchased. If the tax on a $500 item is $30, what is the s
Archy [21]
30 / 500 = x / 900....$ 30 tax / 500 item = $ x tax on 900 item
cross multiply
(500)(x) = (900)(30)
500x = 27000
x = 27000/500
x = 54 <===
6 0
3 years ago
Read 2 more answers
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
5,5,3 Is this a triangle? Explain.
Vlad1618 [11]

Problem 1

Answer: <u>Yes</u> a triangle can be formed with side lengths 5,5,3

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

Explanation:

Take any two sides and add them up. If we get a sum larger than the third side, then a triangle is possible. This is the triangle inequality theorem.

5+5 = 10 is larger than 3

5+3 = 8 is larger than 5

This shows a triangle is possible. The triangle is isosceles because two sides are the same length.

==========================================================

Problem 2

Answer: <u>Yes</u> a triangle can be formed with side lengths 8,8,8

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

Explanation:

This triangle is equilateral because all sides are the same length.

Take any two sides, add them up, and you'll find the sum is larger than the third side.

==========================================================

Problem 3

Answer: <u>No</u> a triangle cannot be formed with sides 7,8,15

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

Explanation:

Yes it is true that

8+15 = 23 is larger than 7

7+15 = 22 is larger than 8

but

7+8 = 15 is not larger than 15

So this means a triangle is not possible.

Imagine you had 3 strips of paper that were 7 inches, 8 inches and 15 inches long. The 7 and 8 inch strips add to 15 inches, but those two smaller strips only form a straight line. We cannot pull them so a triangle forms. If we did, then the third side would be smaller than 15 inches. This is one way to get hands on to see why the triangle inequality theorem works.

==========================================================

Problem 4

Answer: <u>Yes</u> a triangle is possible with side lengths 5,6,10

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

Explanation:

Same idea as before. We can take any two sides, add them, and get a sum larger than the third side. A triangle is possible.

  • 5+6 = 11 is larger than 10
  • 6+10 = 16 is larger than 5
  • 5+10 = 15 is larger than 6
4 0
3 years ago
a rectangular field is 0.45 kilometers long and 0.3 kilometers wide what is the area of the field in square meters?
bekas [8.4K]

Answer:135000

Step-by-step explanation:

5 0
3 years ago
If m∠OZQ = 125 and m∠OZP = 62, what is m∠PZQ
const2013 [10]

Answer:

<h2>               63° or  173°</h2>

Step-by-step explanation:

You didn't specify where the point P is so there are two posibilites:

1.

P₁Z is dividing angle OZQ

then:

      m∠P₁ZQ = m∠OZQ - m∠OZP₁ = 125° - 62° = 63°

2.

P₂Z is outside angle OZQ

then

    m∠P₂ZQ = 360° - (m∠OZQ + m∠OZP₂) = 360° - (125° + 62°) = 173°

5 0
3 years ago
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