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Fittoniya [83]
3 years ago
11

What is 1/(2^3) written as a negative exponent?

Mathematics
2 answers:
fiasKO [112]3 years ago
6 0

Answer:

in negative exponent is 2^-3

Alla [95]3 years ago
6 0

Answer:

in negative exponent is 2^-3

Step-by-step explanation:

in negative exponent is 2^-3

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Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
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\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
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\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
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\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
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\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
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\cfrac{d\theta }{dt}=\cfrac{11}{75}

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We can check each part to make sure. First we start at 4m, then move away, then stay still (zero slope) for three seconds, then move back faster (higher slope). All parts check out ok. 
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