Ask question

Ask question

All categories

3 years ago

8

For science class you have to compare the planets’ distances from the Sun. Because each planet has an elliptical orbit around th

e Sun, it’s distance varies.The chart below shows each planet’s closest and further distances from the sun?

1 answer:

Alchen [17]3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

The length of rectangular field is 7 m less than 4 times the width. The perimeter is 136 m. Find the width and length

Elena-2011 [213]

P = 2(L + W)
P = 136
L = 4W - 7

136 = 2(4W - 7 + W)
136 = 2(5W - 7)
136 = 10W - 14
136 + 14 = 10W
150 = 10W
150/10 = W
15 = W <=== width is 15 m

L = 4W - 7
L = 4(15) - 7
L = 60 - 7
L = 53 <== length is 53 m

3 0

3 years ago

Read 2 more answers

The Bilbo Company plans to manufacture decals for children's lunchboxes. Each decal

Gwar [14]

Answer: 1 -2ax²+2x-ax

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Area of a rectangle: length x width

Replacing with the values given:

A = (2x+1) (1-ax)

A = [2x(1)] + [2x(-ax)] +[ 1(1)]+ [1 (-ax)]

A = 2x- 2ax² +1- ax

A = 1 -2ax²+2x-ax

In conclusion, the correct option is 1 -2ax²+2x-ax

Feel free to ask for more if needed or if you did not understand something.

4 0

3 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

3 years ago

HELP IM CONFUSED PLS PLS WINNER GETS 8 POINTS

MrRa [10]

(10x4x5)+(5x5x5) because you could make 2 shapes out of it. one being a cude that is 5x5x5 and the other being a rectangle that is 4x10x5 thenyou would just add them together. :) hope that helps

3 0

2 years ago

HELP QUICK. By combining three equations, it can be shown that a2+b2=c2. One of the equations is c=e+f. Which of the following c

Alinara [238K]

Answer:

bc=eb and ca=fa

please tell me if this is actually the right answer

Step-by-step explanation:

bc=eb and ac=fa is wrong

bc=eb and ca=fa is right

cb=eb and ac=fa is wrong

ba=eb and ab=fa is wrong

8 0

3 years ago