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soldier1979 [14.2K]
3 years ago
7

∛27x1000????????????????????????????????????

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0

Answer: ∛27x1000= 3x1000 = 3000

Step-by-step explanation:

snow_tiger [21]3 years ago
6 0

Answer:

3000

Step-by-step explanation:

.......... Follow the step above

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PLSSS HURRY ILL GIVE YOU A BRAINLYEST
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Answer:

36 inches

Step-by-step explanation:

lets think of it as a ratio,

80:w

20:9

so we want 20 to get to 80

and 9 to get to w right?

so if we just move the smaller rectangle to the corner, then 20*4=80 right?

and if we do that, we do it to 9 as well, 9*4=36

w=36

6 0
2 years ago
Jose is solving the linear equation 28 - 3x - 15 = -2x + 13 - x. His final two steps are
Alex_Xolod [135]

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A

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6 0
2 years ago
A bag contains 9 red marbles, 18 orange marbles, 2 yellow marbles, and 5 purple marbles. A marble is drawn at random from the ba
vlada-n [284]

Answer:

8/17

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7 0
3 years ago
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galben [10]

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2 years ago
Read 2 more answers
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}
7 0
3 years ago
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