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dexar [7]
3 years ago
8

Plot the inequality x < 0​

Mathematics
1 answer:
Sholpan [36]3 years ago
7 0

Step-by-step explanation:

This is saying " Every number possible thats less than, but not 0" so you show that

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3 0
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3 years ago
Read 2 more answers
How can I solve this?<br><br>x=<br>AD=<br><br>AB= 5X-4<br>DB=x+1​
aliya0001 [1]

Step-by-step explanation:

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3 0
4 years ago
Y=-1/3x+2<br><br> A.<br> B.<br> C.<br> D.
Eva8 [605]

Answer:

C

Step-by-step explanation:

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The y intercept is 2.

7 0
2 years ago
If
likoan [24]

It's easy to show that 7\tan(4x) is strictly increasing on x\in\left[0,\frac\pi8\right]. This means

M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3

and

m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7

Then the integral is bounded by

\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)

\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}

7 0
2 years ago
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