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Inessa05 [86]
2 years ago
5

Quadrilateral ABCD is inscribed in circle P as shown. Which statement is necessarily true? A. `"m"/_A +"m"/_B = "m"/_C + "m"/_D

` B. `"m"/_A + "m"/_C = "m"/_B + "m"/_D ` C. `"m"/_A + "m"/_D = "m"/_B + "m"/_C` D. `"m"/_A + "m"/_B = 2("m"/_C + "m"/_D) `
Mathematics
1 answer:
pogonyaev2 years ago
3 0

Opposite angles of an inscribed quadrilateral are supplementary, so it is true that ...

... B. m∠A + m∠C = m∠B + m∠D . . . . . = 180°

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Find the times (to the nearest hundredth of a second) that the weight is halfway to its maximum negative position over the inter
lesya [120]

Answer:

0.20 and 0.36

Step-by-step explanation:

y(t) = 2 sin (4π t) + 5 cos (4π t)

We wish to convert this to:

y = A sin(ωt + φ)

We know that ω = 4π.  We also know the following:

5 = A sin φ

2 = A cos φ

Divide the first equation by the second equation:

5/2 = tan φ

φ = tan⁻¹(5/2)

Now, square the two equations and add them together.

5² + 2² = (A sin φ)² + (A cos φ)²

29 = A²

A = √29

The equation of the wave is therefore:

y = √29 sin(4π t + tan⁻¹(5/2))

The maximum negative position is -√29.  And half of that is -½√29.

-½√29 = √29 sin(4π t + tan⁻¹(5/2))

-½ = sin(4π t + tan⁻¹(5/2))

7π/6 + 2kπ or 11π/6 + 2kπ = 4π t + tan⁻¹(5/2)

7 + 12k or 11 + 12k = 24t + 6 tan⁻¹(5/2) / π

t = (7 + 12k − 6 tan⁻¹(5/2) / π) / 24 or (11 + 12k − 6 tan⁻¹(5/2) / π) / 24

Trying different integer values of k, we find there are two possible values for t between 0 and 0.5, both when k = 0.

t = (7 − 6 tan⁻¹(5/2) / π) / 24 or (11 − 6 tan⁻¹(5/2) / π) / 24

t ≈ 0.20 or 0.36

6 0
2 years ago
Identify the equation of the circle that has its center at (-27, 120) and passes through the origin.
nasty-shy [4]
<h3><u>The equation of the circle that has its center at (-27, 120) and passes through the origin is</u>:</h3>

(x + 27)^2 + (y - 120)^2 = 15129

<em><u>Solution:</u></em>

<em><u>The equation of a circle is given as:</u></em>

(x-a)^2+(y-b)^2=r^2

Where,

(a, b) is the centre of the circle

r is the radius

We have the centre of the circle (-27, 120)

Therefore,

a = -27

b = 120

Given that, it passes through origin. which means, (x, y) = (0, 0)

Substitute (a, b) = (-27, 120) and (x, y) = (0, 0) in eqn

(0 + 27)^2 + (0 - 120)^2 = r^2\\\\729 + 14400 = r^2\\\\r^2 = 15129

Substitute r^2 = 15129 and (a, b) = (-27, 120) in eqn

(x + 27)^2 + (y - 120)^2 = 15129

Thus the equation of circle is found

6 0
2 years ago
If BD = 7x – 10, BC = 4x – 29, and CD = 5x – 9, find each value.
DerKrebs [107]
BD=7x-10
Answer: x=bd/7 + 10/7

BC=4x-29
Answer: x=Bc/4 + 29/4

CD=5x - 9
Answer: x= cd/5 + 9/5
4 0
2 years ago
Es urgente ¡¡¡ si la MH de a y 4 es 6 y la MH de 8 y b es 12 calcula la MH de a y b
TiliK225 [7]

Answer:

Sabemos que:

MH(x, y) = \frac{2xy}{x+y}

y tenemos que:

MH (a,4) = \frac{2*a*4}{a+4} = \frac{8*a}{a+4} = 6

Con esto podemos encontrar el valor de a:

8*a/(a+ 4) = 6

8*a = 6*(a + 4) = 6*a + 24

8a - 6a = 24

2a = 24

a = 24/2 = 12.

Tambien sabemos que:

MH(8,b) = \frac{2*8*b}{8+b} =\frac{16*b}{8+b} = 12

Y de ahí podemos despejar b:

(16*b)/(b + 8) = 12

16*b = 12*(b + 8) = 12b + 96

16b - 12b = 96

4b = 96

b = 96/4 = 24

Entonces tenemos a = 12 y b = 24, y el MH de a y b es:

MH(12,24) = 2*12*24/(12 + 24) = 24*24/36 = 16

7 0
2 years ago
Clayton Road is 2.25 miles long. Wood Pike Road is 1.8 miles long. Kisha used a quick picture to find the combined length of Cla
tatuchka [14]
I may ask, where is Kisha’s work?
7 0
2 years ago
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