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fredd [130]
3 years ago
9

there are seven people and Luisa's family they have four loaves of cinnamon bread to share equally Luisa wants to know how much

cinnamon bread she will get​
Mathematics
1 answer:
makvit [3.9K]3 years ago
7 0

Answer:

\frac{4}{7} of a loave of bread per person

Step-by-step explanation:

<u>Step 1:  Convert words to expressions </u>

Seven people in Luisa's Family = <em>7 people </em>

Four loaves of cinnamon bread = <em>4 loaves </em>

<u>Step 2:  Divide people from loaves </u>

4 loaves / 7 people  <-  This means they are splitting 4 loaves of bread between 7 people

\frac{4}{7} of a loave of bread per person

Answer:  \frac{4}{7} of a loave of bread per person

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Eddie bought 27 2/5 inches of wire for a home improvement project. How many 12 inches sections can he make from it?
goblinko [34]
Simply divide your wire by 12 and cut off the excess (as it cannot make a full 12-inch section).

27\frac{2}{5}÷12=2\frac{17}{60}

So you have two 12-inch sections and an additional 3.4-inch section.
3 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}

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