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snow_lady [41]
1 year ago
14

PLEASE I NEED HELP ASAP

Mathematics
1 answer:
Annette [7]1 year ago
8 0

The coordinates of the vertex that A maps to after Daniel's reflections are (3, 4) and the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

<h3>How to determine the coordinates of the vertex that A maps to after the two reflections?</h3>

From the given figure, the coordinate of the vertex A is represented as:

A = (-5, 2)

<u>The coordinates of the vertex that A maps to after Daniel's reflections</u>

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A' = (5 - 2, 2)

Evaluate the difference

A' = (3, 2)

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A'' = (3, -2 + 4)

Evaluate the difference

A'' = (3, 4)

Hence, the coordinates of the vertex that A maps to after Daniel's reflections are (3, 4)

<u>The coordinates of the vertex that A maps to after Zachary's reflections</u>

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A' = (-5, -2 + 4)

Evaluate the difference

A' = (-5, 2)

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A'' = (5 - 2, 2)

Evaluate the difference

A'' = (3, 2)

Hence, the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

Read more about reflection at:

brainly.com/question/4289712

#SPJ1

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Step-by-step explanation:

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                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

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                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

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