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3 years ago

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a mechanical engineer designs a robotic rm to be used in an assembly line. one portion of the arm is 5/12 yard long, and the len

gth of the second portion of the arm is 3/12 yard long.What wouldbe the length, in yards, of the robotic arm if the two portions of the arm were measured end to end?

1 answer:

Amanda [17]3 years ago

4 0

Answer:

8/12 yard or 2/3 yard

Step-by-step explanation:

The sum of the two lengths is ...

5/12 yd + 3/12 yd = (5+3)/12 yd = 8/12 yd

This fraction can be reduced by removing a factor of 4 from numerator and denominator.

8/12 yd = 2/3 yd

If the arm portions are end-to-end, the total length is 2/3 yard.

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Please answer this!!

balu736 [363]

Answer:

3,724 cubic inches

Step-by-step explanation:

All you have to do is multiply 19•14•14

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3 years ago

Solve the equation<br>2/3 h - 1/3 h + 11 = 8<br><br>h = ?

Deffense [45]

Answer:

$2 \div 3h - 1 \div 3 h+ 11 = 8 \\ 2 - 1 \div 3h = - 3 \\ 1 \div 3h = - 3 \\ - 9h = 1 \\ h = - 1 \div 9$

-1/9 is the answer

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3 years ago

Simplify: x0 y-3 / x2 y-1

svetlana [45]

$\frac{x {}^{0}y {}^{ - 3} }{x {}^{2}y {}^{ - 1} } \\ = \frac{1y {}^{ - 3} }{ {x}^{2}y {}^{ - 1} } \\ = \frac{1 \times 1}{ {x}^{2} y {}^{2} } \\ \\ = \frac{1}{ {x}^{2}y {}^{2} }$

Step by Step Explanation:

Evaluate the power: Any non-zero expression raised to the power of 0 equals 1
Simplify: Simplify the expression
Calculate the product: Any expression multiplied by 1 remains the same.

<h2>☆彡Hanna</h2>

#CarryOnLearning

8 0

3 years ago

write an equation of the line containing the specified point and perpendicular to indicated line. (-5,6) 4x-y=3

Radda [10]

<h2>Greetings!</h2>

Answer:

y = $\frac{1}{2}x + \frac{29}{4}$

Step-by-step explanation:

First, we need to rearrange the equation to make it y = mx + c

4x - y = 3

y + 3 = 4x

y = 4x -3

So the slope of the line is 4.

The slope of a line perpendicular to an equation is:

$\frac{-1}{gradient}$

So the slope of the perpendicular line is:

$\frac{-1}{4}$ = -0.25

Now to find the equation of a line you can use the following equation:

y - y₁ = m(x - x₁)

Where y₁ is the y-coordinate, x₁ is the x-coordinate and m is the gradient. Plug the values in:

y₁ = 6

x₁ = -5

m = $\frac{-1}{4}$

y - 6 = $\frac{-1}{4}$ (x - - 5)

To get rid of the fraction we need to multiply the whole equation by 4:

4y - 24 = -1(x - - 5)

The two negatives cancel out:

4y - 24 = -1(x + 5)

Multiply the brackets out:

4y - 24 = -x + 5

Now, to rearrange the formula back into y = mx + c

Move the -24 over to the other side making it a +24:

4y = 2x + 5 + 24

4y = 2x + 29

Divide everything by 4:

y = $\frac{2}{4}x + \frac{29}{4}$

y = $\frac{1}{2}x + \frac{29}{4}$

<h2>Hope this helps!</h2>

8 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 > 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

3 years ago