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Ulleksa [173]
3 years ago
8

Line segment BA is tangent to the circle. A circle is shown. Secant D B and tangent B A intersect at point B outside of the circ

le. Secant D B intersects the circle at point C. The length of A B is x, the length of BC is 55, and the length of C D is 120. What is the length of the line segment BA? Round to the nearest unit.
18 units
65 units
88 units
98 units

Mathematics
2 answers:
Mama L [17]3 years ago
4 0

Answer: 88 units

Step-by-step explanation:

KonstantinChe [14]3 years ago
3 0

Answer:

98

Step-by-step explanation:

i just took the test

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The product of twice a number and six is the same as the difference of eleven times the number and 6/5 . Find the number
Rom4ik [11]

Answer:

-6/5

Step-by-step explanation:

solution;

let,the number be x,

product of <u>twice</u><u> </u><u>of</u><u> </u><u>x</u> and <u>six</u> = the <u>difference </u>

<u>difference of eleven times x and 6/5</u>

or, (2x×6)=(11×x)-(6/5)

or, 12x = 11x - 6/5

or, 12x - 11x = -6/5

therefore,x =-6/5

Therefore, the number required is (-6/5) whose product of twice a number and six is the same as the difference of eleven times the number and 6/5 .

3 0
3 years ago
Let y 00 + by0 + 2y = 0 be the equation of a damped vibrating spring with mass m = 1, damping coefficient b &gt; 0, and spring c
stira [4]

Answer:

Step-by-step explanation:

Given that:    

The equation of the damped vibrating spring is y" + by' +2y = 0

(a) To convert this 2nd order equation to a system of two first-order equations;

let y₁ = y

y'₁ = y' = y₂

So;

y'₂ = y"₁ = -2y₁ -by₂

Thus; the system of the two first-order equation is:

y₁' = y₂

y₂' = -2y₁ - by₂

(b)

The eigenvalue of the system in terms of b is:

\left|\begin{array}{cc}- \lambda &1&-2\ & -b- \lambda \end{array}\right|=0

-\lambda(-b - \lambda) + 2 = 0 \ \\ \\\lambda^2 +\lambda b + 2 = 0

\lambda = \dfrac{-b \pm \sqrt{b^2 - 8}}{2}

\lambda_1 = \dfrac{-b + \sqrt{b^2 -8}}{2} ;  \ \lambda _2 = \dfrac{-b - \sqrt{b^2 -8}}{2}

(c)

Suppose b > 2\sqrt{2}, then  λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.

(d)

From λ² + λb + 2 = 0

If b = 3; we get

\lambda^2 + 3\lambda + 2 = 0 \\ \\ (\lambda + 1) ( \lambda + 2 ) = 0\\ \\ \lambda = -1 \ or   \  \lambda = -2 \\ \\

Now, the eigenvector relating to λ = -1 be:

v = \left[\begin{array}{ccc}+1&1\\-2&-2\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}1&1\\0&0\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

Let v₂ = 1, v₁ = -1

v = \left[\begin{array}{c}-1\\1\\\end{array}\right]

Let Eigenvector relating to  λ = -2 be:

m = \left[\begin{array}{ccc}2&1\\-2&-1\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}2&1\\0&0\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

Let m₂ = 1, m₁ = -1/2

m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right]

∴

\left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= C_1 e^{-t}  \left[\begin{array}{c}-1\\1\\\end{array}\right] + C_2e^{-2t}  \left[\begin{array}{c}-1/2\\1\\\end{array}\right]

So as t → ∞

\mathbf{ \left[\begin{array}{c}y_1\\y_2\\\end{array}\right]=  \left[\begin{array}{c}0\\0\\\end{array}\right] \ \  so \ stable \ at \ node \ \infty }

5 0
2 years ago
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