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lianna [129]
3 years ago
15

Giant Department Stores, Inc. purchased bedroom curtains from The Linen Company for $6.19. The markup was $4.20. What was the se

lling price to the nearest cent?
Mathematics
1 answer:
Vesna [10]3 years ago
4 0
When purchased for 6.19, after mark up +4.20 = $10.39
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Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b
trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

7 0
3 years ago
What is the area, in square centimeters, of the trapezoid below? 18.2 cm 9.2 cm 8.3 cm​
zubka84 [21]
Answer:
113.71 cm^2
Explanation m:
Area= a+b/2 h
18.2 + 9.2/2 •8.3 = 113.71
5 0
3 years ago
Put the following equation of a line into slope-intercept form, simplifying all fractions. 4y-4x= -28
V125BC [204]

Answer:

Y= x-7

Step-by-step explanation:

5 0
3 years ago
Help me its due today
Tanzania [10]

Answer:

$65.35

Step-by-step explanation:

  1. First you have to divide 5.4% by 100
  2. Take 5.4/100 and multiply it by 62
  3. Take the number you get from step 2 and add it to 62! (:
4 0
3 years ago
M∠Q = <br> m∠R = <br> m∠S = <br><br> 40 POINTS!!!!!!!
Marizza181 [45]

9514 1404 393

Answer:

  ∠Q = 89°

  ∠R = 123°

  ∠S = 91°

Step-by-step explanation:

It seems easiest to start by finding the measures of each of the arcs. The measure of an arc is double the measure of the inscribed angle it subtends.

  arc QRS = 2·∠P = 114°

So, ...

  arc QR = arc QRS - arc RS = 114° -41° = 73°

The total of the arcs around the circle is 360°, so ...

  arc PQ = 360° -arc PS -arc QRS

  arc PQ = 360° -137° -114° = 109°

__

  ∠Q = (1/2)(arc RS + arc PS) = (1/2)(41° +137°)

  ∠Q = 89°

__

  ∠R = (1/2)(arc PS +arc PQ) = (1/2)(137° +109°)

  ∠R = 123°

__

  ∠S = (1/2)(arc PQ +arc QR) = (1/2)(109° +73°)

  ∠S = 91°

7 0
3 years ago
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