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3 years ago

How far is a 10% of a 2000 kilometer trip

Mathematics

2 answers:

AleksandrR [38]3 years ago

5 0

Answer:

200

Step-by-step explanation:

2000x0.1=200

Dmitry_Shevchenko [17]3 years ago

4 0

Answer:

answer:

10 2000

1x200=

200

Step-by-step explanation:

hope that helps

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Please help me in Geometry!

Masja [62]

Answer:

You get an perpendicular line through given point.

Step-by-step explanation:

In this method of drawing perpendicular line through given point,

you will draw an arc from given point R, so as to cut given line at A and B.

now ,length RA will be equal to length RB (radius of circle)

now, with length RA you will place compass at A and B to cut the arcs at point P (let) on other side of given line.Now, join PA and PB.

The closed RAPB makes a rhombus and diagonal bisect perpendicularly in rhombus.

so you get an perpendicular line through given point.

8 0

3 years ago

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it takes Jupiter 11.9 years to orbit,or go around the sun. Saturn takes 17.6 more years than Jupiter to orbit the sun. About how

pochemuha

Saturn takes 17.6 more years than Jupiter, so 17.6 + 11.9 will get you how long Saturn will take to orbit the sun, which would give you 29.5 years.

8 0

3 years ago

Brand on has 25 buttons.Mason has 6 times as many buttons.How many buttons does Mason have?

Lady_Fox [76]

Answer: Mason has 150 buttons.

Step-by-step explanation: Brandon has 25 buttons. If Mason has 6 times as many, all you have to do is multiply 25 times 6. You end up with 150 buttons.

4 0

3 years ago

Read 2 more answers

PLS HELP! WILL GIVE BRAINLIEST

Leviafan [203]

Answer:

The center (2, 2.5), radius $\sqrt{37}$ / 2

Step-by-step explanation:

The standard form of the equation of a circle is:

( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius

Rewrite the given equation in the standard form:

2x^2 + 2y^2 - 8x + 10y + 2 = 0
x^2 - 4x + y^2 + 5y = -1
x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
(x - 2)^2 + (y + 2.5)^2 = 37/4

The center is:

(2, 2.5)

And radius is:

$\sqrt{37}$ / 2

4 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

2 years ago