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3 years ago

What is the greatest integer that satisfies the inequality below?

Mathematics

1 answer:

BartSMP [9]3 years ago

3 0

Answer:

Step-by-step explanation:

- 2x > 17

x < 17 ÷ ( - 2 )

x < - 8.5

x ∈ ( - ∞ , - 8.5 )

The greatest integer that satisfies the inequality is ( - 9 )

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Solve for x please. :)

aksik [14]

Answer:

x=105

Step-by-step explanation:

75+75+x+x=360

150+2x=360

2x=210

x=105

I hope this is correct and helps you!

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3 years ago

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Four students are competing in a race. In how many different combinations could the four students place in the race?

VMariaS [17]

"Combinations" isn't really the right word, because a combination doesn't take order of elements into account, and the outcome of a race certainly does depend on order. "Permutations" would be the correct term.

If 4 students are competing, then the race has

4! = 4 • 3 • 2 • 1 = 24

possible outcomes.

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2 years ago

Need help please with this question need the answer fast.

AlladinOne [14]

The answer is C, moderate negative association

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3 years ago

13.

dusya [7]

Answer:

B - x=5

Step-by-step explanation:

If y=25 is 250% more than y=10, so multiply 2 by 250% and you get 5

4 0

3 years ago

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

7 0

2 years ago