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Sonja [21]
3 years ago
6

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
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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
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