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3 years ago

What is the difference between graphing the inequality y > x + 5 and y < x + 5

Mathematics

1 answer:

Zepler [3.9K]3 years ago

4 0

Answer:

The graph on the first equation y>x+5 will be shaded above the line. The graph on the second. equation y< x+5 will be shaded below the line.

When y is less than (y<), then the graph is shaded below the line. if y is greater than,( y>) then it will be shaded above the line.

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Given s = 10 and h = 15. Find the total surface area of the square prism.

anastassius [24]

Answer:

150

Step-by-step explanation:

h = hight

s = sides

surface = 10 x 15

surface = 150

(V = 10 x 10 x 15

(V = 100 x 15)

(V = 1500)

(Base area = B)

(B = Lenght x Width)

(B = 10 x 10)

(B = 100)

3 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

3 years ago

Help me plsssssssssssss

LUCKY_DIMON [66]

Answer:

The radius is 5 cm

Step-by-step explanation:

The formula for the area of a circle is : $\p*radius^2$

To work out the length of the radius you would first need to divide the area of 78.5 cm by pi (pi=3.14), this gives you $25cm^2\\$ . This is because by dividing the area by pi we are isolating the value of the radius squared.

The final step is to work out the radius. You can do this by finding the square root of the radius squared which is 25, this gives you 5 cm. This means that the radius is 5 cm. This is because the square root is a number that when squared equals that number.

1) Divide 78.5 by pi.

$78.5/\pi=25cm^2$