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3 years ago

Please help me with my math

Mathematics

1 answer:

jeyben [28]3 years ago

3 0

25.3 is what i got for the first image.

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Please answer I will give a good review:)

marin [14]

Answer:

Step-by-step explanation:

(8+6x6)= 44

44/11=4

7x2=14

14+4=18

hope this helped

8 0

3 years ago

Read 2 more answers

A running track consists of two parallel lines that are connected at each end but the curved boundary of a semicircle. The paral

saveliy_v [14]

Answer:

The area inside the running track is 248.5 m².

Step-by-step explanation:

The area inside the running track is given by the sum of the area of a rectangle and the area of a semicircle:

$A_{t} = A_{r} + 2A_{c}$

$A_{t} = x*(2r) + 2*(\frac{\pi r^{2}}{2})$

Where:

x: is one side of the rectangle = 30 m

r: is the radius = half of the other side of the rectangle = 7/2 m = 3.5 m

Hence, the total area is:

$A_{t} = 30*7 + \pi (3.5)^{2} = 248.5 m^{2}$

Therefore, the area inside the running track is 248.5 m².

I hope it helps you!

3 0

3 years ago

What is the sum (2 – 3i) + (-4-2i) ?

zysi [14]

-2-5i is the correct answer

5 0

2 years ago

Write the slope intercept form through (3, 5) parallel to y 5/3x -5

Harrizon [31]

Answer:

y=5/3x .This is the answer

8 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