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Nadya [2.5K]
3 years ago
6

What’s the line of best fit ?

Mathematics
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

its the line that goes through a set of points that most accurately shows the relationship between the points.

Step-by-step explanation:

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Question 3 (2 points) Saved Which of the following has the larger area? A rectangle with sides 12 inches and 15 inches. A triang
ZanzabumX [31]

Answer:

rectangle = 180

triangle = 350

the triangle has a much larger area.

350-180=170

the triangle by 17

square inches

5 0
3 years ago
.002 is 1/10 of what
amid [387]
Let the unknown number be x.  Then (1/10)x = 0.002.  To solve for x, multiply both sides by 1000 to remove the decimal fraction 0.002.

Then 1000(1/10)x = 1000(0.002), or 100x = 2.  Dividing both sides by x returns 

x=100/2, or x= 50 (answer)
5 0
3 years ago
Write an equation that illustrates this function if you plan to keep the truck for 7 years and determine the value of the
lisov135 [29]

Answer:

well, this answer is considered unavailable equals. this means the question is being asked wrong. none of the answers are correct as there is no way to make an available answer from the statement.

6 0
3 years ago
(a) Find the first 3 terms in ascending powers of x of the binomial expansion of (2+x/2)^6
Firdavs [7]

Answer:(2+x/2)^6=64(1+3x/2+15x^2/16+5x^3/16+....) (2.05)^6=74.2203

Step-by-step explanation:

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
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