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3 years ago

Find x and round to the nearest tenth:)

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer:

x = 31.8°

Step-by-step explanation:

Applying the Law of Sines, we have:

$\frac{sin(x)}{12} = \frac{sin(75)}{22}$

Multiply both sides by 12

$\frac{sin(x)}{12} \times 12 = \frac{sin(75)}{22} \times 12$

$sin(x) = \frac{sin(75) \times 12}{22}$

$sin(x) = 0.5269$

$x = sin^{-1}(0.5269)$

x = 31.8° (nearest tenth)

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Step-by-step explanation:

The given graph is a transformed logarithm function.

The graph is obtained by shifting the parent function three units left.

The vertical asymptote is now

$x = - 3$

The x-intercept is where the graph intersect the x-axis, which is x=-2

Therefore the last option is correct.

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The local minima of $f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$ are (x, f(x)) = (-1.5, 0) and (7.980, 609.174)

<h3>How to determine the local minima?</h3>

The function is given as:

$f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

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(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

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3 years ago

Let y 00 + by0 + 2y = 0 be the equation of a damped vibrating spring with mass m = 1, damping coefficient b > 0, and spring c

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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

3 years ago