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Sedbober [7]
2 years ago
6

29

Mathematics
1 answer:
Lelu [443]2 years ago
7 0

Answer:

b easy question but can I get brainliest

Step-by-step explanation:

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What is 36/40 as a percentage and can you show me how to work it out as well please thanks x
cricket20 [7]
36/40 = 90%
<span>
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36/40= 0.9
0.9x100 =90
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6 0
2 years ago
Read 2 more answers
Help help help help help
Naddik [55]

Answer:

<, -4

Step-by-step explanation:

-8x-20<12

add 20 to both sides

-8x=32

divide both sides by -8

x=-4

3 0
2 years ago
If T(x, y) = (x + 5, y + 6) and Pris the image of P, what is the rule for the translation in which P is the image of P'? T(x, y)
xxTIMURxx [149]

Answer:

The translation is;

T(x,y)=(x-5,y-6)

Explanation:

Given the translation rule;

T(x,y)=(x+5,y+6)

when P' is the image of P.

For the inverse, when P is the image of P', the Translation rule would become;

\begin{gathered} x=x^{\prime}+5 \\ x^{\prime}=x-5 \\ y=y^{\prime}+6 \\ y^{\prime}=y-6 \\ So,\text{ the translation rule becomes;} \\ T(x,y)=(x-5,y-6) \end{gathered}

The translation is;

T(x,y)=(x-5,y-6)

3 0
1 year ago
Explain how you can use division to simplify a complex fraction.
PolarNik [594]
Reduce the final answer if needed (or reduce when you multiply).  A mixed number should be converted to an improper fraction (heavier on top) before simplifying. Change the numerator and denominator to single fractions<span> by creating a common denominator</span>
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2
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
1 year ago
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