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Nimfa-mama [501]
3 years ago
15

Suppose that a car normally sells for $16,500 and now on sale ft $14,000. What is the percent of discount? If necessary round yo

ur answer the nearest tenth.
Mathematics
1 answer:
krek1111 [17]3 years ago
5 0
Percentage increase/decrease is given by (the difference ÷ original value) × 100

The decrease in value is 16500 - 14000 = 2500

The percentage decrease = (2500÷16500) × 100 = 15.2%
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zysi [14]
<h3>Answer:  0 \le \theta < \frac{\pi}{4}</h3>

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

How to get this answer:

Use the unit circle to note that \sin\theta = \cos\theta = \frac{\sqrt{2}}{2} when \theta = \frac{\pi}{4} (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point \left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)

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

Here's a more detailed picture of whats going on.

\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\

Intersect the intervals 0 \le \theta < \pi and \theta < \frac{\pi}{4} and you'll end up with the final answer 0 \le \theta < \frac{\pi}{4}

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3 years ago
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Answer:

9.9.9.9 = ( 9*1) . (9/10) + (0/100) + (9/1000) + (0/10000) +(9/100000)

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3 years ago
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