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Anettt [7]
3 years ago
15

Help somebody? Please

Mathematics
1 answer:
Alja [10]3 years ago
6 0

Answer:

I think you have the right answer. Since it is only -3 the line would not go into the positives and it says nothing about being positive, so I would say the answer is the one that goes from -3 to 0. I'm not sure how to show the steps but I explained my answer to this problem so it counts as showing my work. Thank you! Hope this helps! :)

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David is a groundskeeper at Triangle Park, scale shown below.
stellarik [79]
Oh you got an answer good luck
3 0
2 years ago
Estimate the value 9.03 + 19.87 x 3.11 - 4.97
dybincka [34]

Answer:

82

Step-by-step explanation:

For this problem you would first round everything. 9.03 becomes 9, 19.87 becomes 20, 3.11 becomes 3 and 4.97 becomes 5. You then just do the problem. 9 + 20 = 29, multiplied by 3 makes 87, 87 - 5 = 82.

8 0
3 years ago
Everbank Field, home of the Jacksonville Jaguars, is capable of seating 76,867 fans. The revenue for a particular game can be mo
Andrei [34K]
The domain of the function is all the values from which the functio can be mapped: here it's between 0 and <span>76,867 -depending on how many people come

</span>0 \leq x\geq 76867

The range is all the values that the function can have.So here it's from 0, when noone is coming, to 76867*161=12377036

0 \leq y\geq 12377036




5 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
. How many faces does a polyhedron with<br> 20 vertices and 30 edges have?
Alisiya [41]

Answer:

12

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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