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Daniel D.
Princeton Grad teaching math, science, and anything else
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TutorMe
Question:

In the United States, income taxed in brackets, so that the first $10,000 a person makes are taxed at 10%, the next $30,000 a person makes are taxed at 15%, and the next $50,000 after that are taxed at 25%. Which of the following expressions accurately describes the taxes ($$T$$) paid by a person with an income ($$I$$) greater than $50,000 and less than $90,000? (a) $$T = 0.25 I$$ (b) $$T = 0.10 (I - 10000) + 0.15 (I - 30000) + 0.25 (I - 50000) $$ (c) $$T = 0.10 (10000) + 0.15 (30000) + 0.25 (I - 40000)$$ (d) $$T = 0.10 (I - 40000) + 0.15(I - 20000) + 0.25I$$

Daniel D.
Answer:

The answer is (c). (a) is wrong because only the last part of the income is taxed at 25%, not the whole income. (b) and (d) are wrong because if the income is greater than $50,000, more than $10,000 will be taxed at the lowest bracket, and more than $30,000 will be taxed in the middle bracket. Only (c) matches what was said in the problem, that up to 10,000 is taxed at 10%, and no more, and up to 30,000 is taxed at 15%, and no more, and the rest past that 10,000 and 30,000 will be taxed at 25%.

Calculus
TutorMe
Question:

What is the derivative with respect to $$x$$ of $$\cos(2^{2x})$$?

Daniel D.
Answer:

We have to apply three rules in order from the outside in: (1) the derivative with respect to $$x$$ of a cosine of something is equal to the negative sine of the something times the derivative of the something with respect to $$x$$ (that is, $$\frac{d}{dx} \cos u = -\sin u \frac{d}{dx}u$$), (2) the derivative with respect to $$x$$ of an exponential is equal to the exponential times the natural logarithm of the base times the derivative of the exponent with respect to $$x$$ (that is, $$\frac{d}{dx} n^u = n^u \ln n \frac{d}{dx}u$$), and (3) the derivative with respect to $$x$$ of $$nx$$ is $$n$$. Applying the rules, we get: First: $$\frac{d}{dx} \cos(2^{2x}) = -\sin(2^{2x}) \frac{d}{dx} (2^{2x})$$; second, $$ -\sin(2^{2x}) \frac{d}{dx} (2^{2x}) = (-\sin(2^{2x}))(2^{2x})(\ln 2) \frac{d}{dx} 2x$$; and third, $$ (-\sin(2^{2x}))(2^{2x})(\ln 2) \frac{d}{dx} 2x = (-\sin(2^{2x}))(2^{2x})(2 \ln 2)$$

Physics
TutorMe
Question:

A driver is accused of having been speeding before colliding with another car that broke down on the interstate. Police at the scene of the accident measure tire marks that extend from the point that the driver began braking to the point of impact, 80 $$m$$ apart. The brakes on the driver's car are capable of decelerating the car up to 8 $$m/s^2$$. The speed limit on the road is 30 $$m/s$$. Was the driver speeding?

Daniel D.
Answer:

Yes. We can use the equation $$v^2 = v_0^2 + 2ax$$ to determine the initial velocity of the car ($$v_0$$), given the other values. Plugging in $$v = 0 m/s$$, $$a = -8 m/s^2$$, and $$x = 80 m$$, we get $$0 = v_0^2 + 2(-8)(80) m^2/s^2$$, so $$v_0^2 = 1280m^2/s^2$$, and $$v_0 \approx 35.7 m/s$$. The driver was speeding.

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