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3 years ago

Domain of the function

Mathematics

1 answer:

dimaraw [331]3 years ago

4 0

Hello :
y exist : x+6 <span>≥ 0
x </span>≥ -6

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At the beginning of my bike ride I feel good, so I can travel 20 miles per hour. Later, I get tired and travel only 12 miles per

Vesnalui [34]

Answer: $3\dfrac{1}{4}$ hours

Step-by-step explanation:

Let x be the time ( in hours ) you feel good and y be the time ( in hours ) that you get tired.

As per given , we have

x+y= 8

Since : $Distance = Speed\times Time$

So , we have

20x+12y=122

According to the given question , we have the following system of equations ,

$x+y= 8--------(1)\\\\20x+12y=122------------------(2)$

Multiply 12 on both sides of equation (1) , we get

$12x+12y=96------------(3)$

Eliminate equation (3) from (2) , we get

$8x=26\\\\\Rightarrow\ x=\dfrac{13}{4}=3\dfrac{1}{4}$

Hence, you felt $3\dfrac{1}{4}$ hours good.

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: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

PLZ HELP IM IN DESPERATE NEED OF AN ANSWER!!!

Irina18 [472]

I believe 4(x - 3)=32 could help I haven’t done an equation from the unit in months so hopefully that looks familiar to you.

5 0

2 years ago

What is the product?<br> I need help

LenaWriter [7]

Answer:

honey there is no picture to use...

Step-by-step explanation:

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3 years ago

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George finished 10 of 25 math problems.what percent of the problems did george finish

PIT_PIT [208]

Answer:

40%

Step-by-step explanation:

So 10/25=0.4

0.4=40%

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3 years ago

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