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3 years ago

Cual es el opuesto de -12, 63 y -47 ¿?

Mathematics

1 answer:

scoundrel [369]3 years ago

3 0

El opusesto sera lo contrario,si es negativo de convertira a postivo y si es positivo de convierte ah negativo.

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Can segments with lengths of 15, 20, and 36 form a triangle?

strojnjashka [21]

Ok, so remember

the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle

so the longest side is 36

36 must be less than the sum of 15 and 20
36<15+20
36<35
false

therfor it cannot form a triangle
try it yourself, you cannot connect them that way

5 0

3 years ago

A jet flies over the Air Force Base at 10:20 a.m. At 10:32 a.m., the plane passes over the Navy Base Station, 120 miles away. Ho

kati45 [8]

So first calculate the time between the plane flied over the Air Force Bases (AFB) and Navy Base Station (NBS):
(time passed Air Force Bases) - (time passed Air Force Bases) = 10:32 - 10:20 = 12 mins
Since we know the distance that plane travel from AFB to NBS is 120 miles and the plane traveled that distance within 12 minutes.
We have
120 miles / 12 minutes = 10 miles / minute
Calculate it into miles per hour (60 minutes):
10 miles/minute x 60 minutes = 600 miles per hour.
ANSWER : 600 mph

3 0

3 years ago

Is anyone good at geometry???

zhenek [66]

Hey there Smarty!

This would be considered to be a (acute angel) which in this case, we would have to make sure that this whole triangle would equal less than 270 because each angle would be less than $90$ °

If we add/multiply $\boxed{(90+90+90) \ or \ (90*3) = 270}$ .

So, we would have to know that was ever this would all add up to be, this would have to be less than 270°

$\left[\begin{array}{ccc}\boxed{\boxed{(2.5+5) \\ \\ (2.5*2.5) \\ (2.5-2) \\ (2.5-1) \\}} \\ \\ this \ would \ be \ why \ I \ would \ say \\ that \ this \ would \ be \ the \ answer \end{array}\right]$

I truly hope this helps, and also, it's kind of my
first time doing this I hope this would be helpful.
~Jurgen

7 0

3 years ago

Charlie bought a pair of shorts at the store when they were having a 45% off sale.If the regular price of the pair of shorts was

nekit [7.7K]

Charlie would have to pay 10 dollars

7 0

3 years ago

Read 2 more answers

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