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garik1379 [7]
3 years ago
8

How many hours are in 3 years

Mathematics
2 answers:
777dan777 [17]3 years ago
7 0
26280 hours are in 3 years.
svp [43]3 years ago
4 0
Hello,

shall we begin?

Years = 3
Year = 365 days
1 day = 24 hours


= 3 * 365 * 24
= 26.280

Answer: 26,280 hours



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Suppose the sum of two integers is negative. Must both integers be negative? Explain.
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Not necessarily. There are many ways to write a basic equation with a negative answer. For example, -3-4 = -7. Here 4 is a positive number but because you subtract 4 to 3 or take away 4 from 3 being a negative number and so you get a negative answer. Another example is 6+(-9). There are a couple of ways you can resolve this. My method is to subtract 9 from 6 which gives you 3 and simply add a negative sign.

Let me know if you'd  like more examples. Hope this helps!! Sorry if it is confusing I can explain you in a more simpler way if you'd like.

Step-by-step explanation:

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Assume that the radius of the circle is 1.


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We can conclude that the white area is about 35%.

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What number is 7 units to the left of -1?<br><br> a.6<br> b.-6<br> c.-8<br> d.8
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... c. -8

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Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
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}
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