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2 years ago

4X9 help plsss I will give branilyst

Mathematics

2 answers:

Feliz [49]2 years ago

4 0

Answer:

Step-by-step explanation:

9+9=18+18=36

jenyasd209 [6]2 years ago

3 0

Answer:

Step-by-step explanation:

9+9+9+9=36

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BRAINLIEST !!!PLEASE HELP

cricket20 [7]

Marcella would not be able to win the competition because her kite string (159.75 feet) is less than 162 feet.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Trigonometric ratio is used to show the relationship between the sides and angles of a right angled triangle.

Let l represent the length of the string, hence:

sin(28) = 75/l

l = 159.75 feet

Marcella would not be able to win the competition because her kite string (159.75 feet) is less than 162 feet.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

6 0

2 years ago

In an arithmetic sequence, if a2= 5 and a10= 19 find a6

OLEGan [10]

Answer:

a₆ = 12

Step-by-step explanation:

The nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₂ = 5 and a₁₀ = 19 , then

a₁ + d = 5 → (1)

a₁ + 9d = 19 → (2)

Subtract (1) from (2) term by term to eliminate a₁

8d = 14 ( divide both sides by 8 )

d = 1.75

Substitute d = 1.75 into (1)

a₁ + 1.75 = 5 ( subtract 1.75 from both sides )

a₁ = 3.25

Then

a₆ = 3.25 + 5(1.75) = 3.25 + 8.75 = 12

6 0

2 years ago

3x squared - 2 x y when x=2 and y = 5

ira [324]

the Answer is 4 do order of operations

7 0

3 years ago

9 out of 20 students will attend the Worldwide Day of Play at RMS what percentage of students will not attend

EastWind [94]

U do 4/20 = p/100 then u cross multiply the 4 with 100 getting u 400, then divide the 400 by 20, giving u the answer of 20%

8 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