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motikmotik
3 years ago
9

5/16 divide by 20/40= SHOW WORK

Mathematics
2 answers:
Elenna [48]3 years ago
6 0
5/16 divided by 20/40
5/16 times 40/20
5 times 40= 200
16 times 20=320
200/320=20/32=10/16=5/8

Final Answer: 5/8
Tanya [424]3 years ago
3 0
5/16 ÷ 20/40 = 5/16 × 40/20 = 10/16 = 5/8
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Answer:

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HURRY. NEED To be done by 10:30
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Brandon sights a helicopter above a building that is 200 feet away at an angle of elevation of 30 degrees. To the nearest foot,
Firlakuza [10]
The problem says that <span>Brandon sights a helicopter above a building that is 200 feet away at an angle of elevation of 30 degrees. So, you can calculate the height asked, by following this procedure:
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 Tan(α)=Opposite leg/Adjacent leg

 α=30°
 Opposite leg=x
 Adjacent leg=200 feet

 When you substitute these values into the formula above (Tan(α)=Opposite leg/Adjacent leg), you have:

 Tan(α)=Opposite leg/Adjacent leg
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4 0
3 years ago
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vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
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