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tiny-mole [99]
3 years ago
12

The coordinates of point P on a coordinate grid are (−5, −6). Point P is reflected across the y-axis to obtain point Q and acros

s the x-axis to obtain point R. What are the coordinates of points Q and R?
1. Q(5, 6) and R(−5, −6)

2. Q(−5, −6) and R(5, 6)

3. Q(−5, 6) and R(5, −6)

4. Q(5, −6) and R(−5, 6)
Mathematics
2 answers:
Natali [406]3 years ago
7 0
Q(−5, 6) and R(5, −6) hope I helped you :)
Llana [10]3 years ago
4 0
Hey friend!
Let's figure this out!

P(−5, −6)
reflect about y axis
Q(5, -6)
reflect about x axis
R(5,6)


So that gives you the answer!
3. Q(−5, 6) and R(5, −6)



Hope this helped!
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Which property is used in the first step to solve this equation?
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Answer:

distributive property

Step-by-step explanation:

we know the distributive property states that the multiplication of a number (in this case, the number 4) by a sum (in this case is a subtraction:  x-6) is equal to the sum of the multiplications of that number for each of the terms that belong to the sum:

we have: 4(x-6)

we apply the distributive property, so we have

4(x-6)=4*x-4*6

4(x-6)=4x-24

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3 years ago
A line is to be graphed through the point (–3, 0). Which points can be used to create a line with an undefined slope through (–3
irina1246 [14]

Answer:-3.0

Step-by-step explanation:

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4 years ago
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Does 125/35 and 25/7 form a proportion
White raven [17]

Answer:

Yes it does

Step-by-step explanation:

(125/35)=(25/7)

((125/5)/(35/5))=(25/7)

(25/7)=(25/7)

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}

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Here s the question this is the doc i put it on please help
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