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2 years ago

Which of the following is a segment congruent to GJ¯¯¯¯¯ ?

Mathematics

1 answer:

LiRa [457]2 years ago

7 0

Answer:

C. KN

Step-by-step explanation:

In the line you can see GJ is 3 units and KN is the only answer to be 3 units apart as well. So your answer is C!

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I don’t get 15a. The t^5 is confusing me

kondor19780726 [428]

15a is non-exponential because the variable is not in the exponent

Instead, this is a polynomial function. It can also be called a power function as power functions are in the form a*x^b where 'a' and 'b' are fixed numbers.

3 0

3 years ago

If f(1) = 160 and f(n + 1) = –2f(n), what is f(4)?

sashaice [31]

Hello,

f(1)=160
f(2)=160*(-2)
f(3)=160*(-2)²
f(4)=160*(-2)^3=-1280

f(n)=160*(-2)^(n-1)

8 0

3 years ago

Read 2 more answers

Pizza Perfection sold 93 pieces on Monday 67 pizzas and Tuesday 78 pizzas on Wednesday 44 piece pizzas on Thursday and 123 pizza

Irina18 [472]

93+67+78+44+123/5= 81 the answer

8 0

3 years ago

3/4(12x−8)+3=10−1/2(6x+2)

Aleonysh [2.5K]

Answer:

$x=1$

Step-by-step explanation:

$\frac{3}{4}(12x-8)+3=10-\frac{1}{2}(6x+2)$

$(9x-6)+3=10-(3x+1)$

$9x-6+3=10-3x-1$

$9x-3=9-3x$

$9x=12-3x$

$12x=12$

$x=1$

8 0

2 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