Step-by-step explanation:
g(x)= -16x² + 72x + 80
Plug 2 in to check if it's correct.
g(2)= -16(2)² + 72(2) + 80
-16(4)+128+80
-72+128+80
= 136
Answer:
Discount = $290.76 × 0.2 = $58.15
Cost After Discount = $290.76 - $58.15 = $232.61
Sales Tax = 0.0625 × $232.61 = $14.56
How Much You Pay = $232.61 + $14.56 = $247.17
Step-by-step explanation:
Discount:
290.76 × 0.2* = 58.152
Round that to 58.15
* Turn 20% into a decimal (0.2)
Cost after Discount:
290.76 - 58.15 = 232.61
Sales Tax:
First turn 6.25 % into a decimal. Which would be 0.0625.
0.0625 × 232.61* = 14.556875
Then we round to the nearest hundredth and you get 14.56.
* We us the amount after the discount instead of the original price.
How much you pay:
232.61 + 14.56 = 247.17
5 is in the ten thousands place now. Since the next digit (a 6) is greater than 5, we must round UP: 360000 is to the nearest ten thousand.
The product of a non-zero rational number and an irrational number is irrational.
Therefore C is the answer.
A, B, and D are all properties of rational numbers. Irrational numbers are ones that can not be written as a ratio, or fraction, using whole numbers. They are non-repeating, non-terminating numbers. Example would be pi.
Answer: 
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Work Shown:
Apply the following trig identity

Now let's rationalize the denominator

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As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.
Or you can note how

which helps us see that they are the same thing.
Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as
but it's the same as above.