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Darina [25.2K]
4 years ago
13

Trig question -- please help if you can!

Mathematics
1 answer:
yanalaym [24]4 years ago
4 0
\bf \textit{Sum and Difference Identities}
\\\\
sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)
\\\\
cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\\\\
-------------------------------\\\\
y=r[cos(\theta +R)]\implies y=r[cos(\theta )cos(R)-sin(\theta )sin(R)]
\\\\\\
y=rcos(\theta )cos(R)-rsin(\theta )sin(R)
\\\\\\
z=r[sin(\theta +R)]\implies y=r[sin(\theta )cos(R)+cos(\theta )sin(R)]
\\\\\\
z=rsin(\theta )cos(R)+rcos(\theta )sin(R)
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Please help what is the answer
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Answer: D

Step-by-step explanation:

The degree of the equation is 1.

The reason it's not A is because |x| can be represented as a piecewise function composed of 2 different linear functions.

7 0
3 years ago
What is the volume of a rectangular prism that has a length, width, and height of 12 m, 14 m, and 23 m, respectively?
AveGali [126]
Find the volume with this formula
v = l × w × h

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l = 12 m
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6 0
4 years ago
1000 divided by1000 is just 1
ladessa [460]

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Step-by-step explanation:

5 0
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6 0
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Am i correct? calculus
Advocard [28]
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
\\\\\\
\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}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\


\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\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}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}

8 0
3 years ago
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