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bagirrra123 [75]
3 years ago
7

The rectangle ABCD is reflected across the y-axis. What are the coordinates of the vertex A'?

Mathematics
1 answer:
Anit [1.1K]3 years ago
5 0
The answer would be (-2,5), this is because when we reflect on the Y axis the Y stays the same and X changes and vise versa for the X axis. Then the amount of units away from the y axis is the amount of units more that it’s going to travel. Hope this helps
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The total charge on 7 particles is −42 units. All the particles have the same charge.
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Read 2 more answers
Hey what is X+y please help
dusya [7]

Answer:

X = 2
Y = 8

Step-by-step explanation:

X is 2 because half of 4 is 2 and Y is 8 because double of 4 is 8.

4 0
2 years ago
(HURRY! I'M BEING TIMED)Write the partial fraction decomposition of the rational expression.
Aleonysh [2.5K]

Answer:

The partial fraction decomposition is \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}.

Step-by-step explanation:

Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions.

To find the partial fraction decomposition of \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}:

First, the form of the partial fraction decomposition is

                                  \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x + 2}

Write the right-hand side as a single fraction:

                             \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B}{\left(x + 1\right)^{2} \left(x + 2\right)}

The denominators are equal, so we require the equality of the numerators:

             - 4 x^{2} + 13 x - 12=\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B

Expand the right-hand side:

           - 4 x^{2} + 13 x - 12=x^{2} A + x^{2} C + 3 x A + x B + 2 x C + 2 A + 2 B + C

The coefficients near the like terms should be equal, so the following system is obtained:

\begin{cases} A + C = -4\\3 A + B + 2 C = 13\\2 A + 2 B + C = -12 \end{cases}

Solving this system, we get that A=50, B=-29, C=-54.

Therefore,

                                  \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}

7 0
3 years ago
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