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3 years ago

Frank hikes 2.25 miles in 1/2 an hour. What is his rate in miles per hour?

Mathematics

1 answer:

Valentin [98]3 years ago

5 0

4.50 MPH

because 2.25 times 2 is 4.50

and it is 1/2 an hour

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What is 10% of 45.00

Tems11 [23]

Answer:

4.5

Step-by-step explanation:

$\frac{10x45}{100}$ =4.5

5 0

3 years ago

Read 2 more answers

Find the distance along an arc on the surface of the earth that subtends a central angle of 6 minutes (1 minute = 1/60 degree).

bazaltina [42]

Answer:

6.914miles (to 3 decimal places)

Step-by-step explanation:

Length of an arc = θ/360 *2πr

angle subtended by arc at the center of the earth = 6 minutes

1 minute =1/60 degree

6 minute= 6/60 degree =0.1°

θ=0.1°

radius of earth = 3960 miles

distance along the arc = $\frac{0.1}{360}*2\pi *3960$

=6.9142857143miles

3 0

2 years ago

An MP3 player costs $70 and song downloads cost $0.95 each. Write an expression that represents the cost of the MP3 player and x

babunello [35]

Let x represent the number of song downloads and let y represent the total cost

y=0.95x+70

3 0

3 years ago

What is the approximate volume of the small truck? length 11 1/3 ,width 7 5/12 ,height 6 3/4?

luda_lava [24]

V = L X W X H
V = 11 1/3 X 7 5/12 X 6 3/4
V = 567 54/144
Which reduces to 567 3/8
V= 567 3/8

8 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