All categories

2 years ago

HELP ASAP! What is the slope of the line???

Mathematics

1 answer:

ANEK [815]2 years ago

8 0

Answer: the slope is 1

Step-by-step explanation:

Slope is basically rise over run. Looking at the graph you see it rises one unit and then goes over to the left by one unit. 1÷1 equals 1 so That is your slope.

You might be interested in

Help fast I'm dum

Llana [10]

Answer:

384

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

£38.40 divided by 16

Nikitich [7]

It is 2.4 because 16 goes twice and the remainder is 4

4 0

3 years ago

Read 2 more answers

the area of aron's favorite poster is 6 square feet. the length of the poster is 3 feet. find the width of the poster

MatroZZZ [7]

Answer:

The width is 2 ft

Step-by-step explanation:

We know the area of a rectangle is

A = l*w

The area is 6 and the length is 3

6 = 3*w

Divide each side by 3

6/3 = 3w/3

2 =w

The width is 2 ft

6 0

2 years ago

Read 2 more answers

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

Please help me with the question. I need help with the top question not the bottom

Schach [20]

Yes-this is because you can start from the 7th part and then go backwards to forwards to find the letter/number/pattern you need

3 0

2 years ago