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3 years ago

The moon is about 240,000 miles from earth what is the order of magnitude of this number

2 answers:

5 0

The order of magnitude is calculated by use of log function. log(240000) = 5.38 so the order is 5 (from what I've see we round down even if it was 5.999 but this may be wrong in some cases).

olga55 [171]3 years ago

4 0

<span>1578620 this is what i got i hope it helps you

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"Nine more than three times a number is negative six"<br><br>what is the answer?

vekshin1

I'm sorry I have No clue

Step-by-step explanation:

lol but I honestly d0n7 kno

3 0

2 years ago

Guys please help me with this question

Kruka [31]

Answer:

if it's gonna equal 12 then a and b equals 6

Step-by-step explanation:

a/b

a=6

b=6

6/6

4 0

2 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

2 years ago

Clear parentheses and combine like terms:

mihalych1998 [28]

Answer:

-3p - p²

Last option

Step-by-step explanation:

Step 1: Write out expression

2p - 7p² - 5p + 6p²

Step 2: Combine like terms (p²)

-p² + 2p - 5p

Step 3: Combine like terms (p)

-p² - 3p

Step 4: Rewrite

-3p - p²

7 0

3 years ago

Can someone help me please

Oxana [17]

C

3 3/4 = 3.75
21/6 = 3.5

5 0

2 years ago