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3 years ago

How is a 15% discount similar to a 15% decrease? Explain.

2 answers:

andrey2020 [161]3 years ago

7 0

They are the same. Example: If something costs 100$ and you found a coupon to get a 15% discount you will save: 15$ and your cost will be 100$-15$=85$ Hope this helped! Have a great day!

Furkat [3]3 years ago

7 0

15percent discount means they put the price lower by 15percent and 15percent decrease is also low down the price by 15percent..doesn't that make both means lower the price by 15 percent..

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How do you find the y - intercept (explain in sentences)

Aloiza [94]

Find the x-intercept of a given linear equation, plug in 0 for 'y' and solve for 'x'. To find the y-intercept, plug 0 in for 'x' and solve for 'y'. In this tutorial, you'll see how to find the x-intercept and the y-intercept for a given linear equation. Check it out!

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3 years ago

Is 0.75 a solution for n < 1

attashe74 [19]

Yes it could be one........

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3 years ago

12 + 2(x + 6) - 8 = 26

Kisachek [45]

Answer: x=5

Step-by-step e

6 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

which statement is true about the equations –3x 4y = 12 and x1/4 – 1/3y = 1? the system of the equations has exactly one solutio

finlep [7]

If you would like to solve the system of equations, you can do this using the following steps:

-3x + 4y = 12
x * 1/4 - 1/3 * y = 1 ... x * 1/4 = 1 + 1/3 * y ... x = 4 + 4/3 * y
_____________
<span>-3x + 4y = 12
</span>-3 * (4 + 4/3 * y) + 4y = 12
-12 - 4y + 4y = 12
-12 = 12
-12 - 12 = 0
-24 = 0

The correct result would be: <span>the system of the equations has no solution; the two lines are parallel.</span>

4 0

3 years ago