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Rashid [163]
3 years ago
5

The vertices of a triangle are P(2, –8), Q(1, –4), and R(4, –1). Name the vertices of P'Q'R' after a reflection over the x-axis.

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

The coordinates of P'Q'R' are: P' (2,8), Q' (1,4) ,R'(4,1)

Step-by-step explanation:

We are given:

The vertices of a triangle are P(2, –8), Q(1, –4), and R(4, –1).

We need to:

Name the vertices of P'Q'R' after a reflection over the x-axis.

Reflection over the x-axis

The rule used for reflection over x-axis

P(x,y)= P'(x,-y)

The coordinates of x-axis remains same, while coordinates of y-axis become negative.

Now, finding coordinates of P'Q'R'

P (2,-8) -> P' (2,8)

Q(1,-4) -> Q' (1,4)

R(4,-1) -> R'(4,1)

So, the coordinates of P'Q'R' are: P' (2,8), Q' (1,4) ,R'(4,1)

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Answer:

-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14 : x = \frac{840}{180^{\circ \:}}

Decimal Form:

x = -267.38030...

Step-by-step explanation:

-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14

Remove Parentheses: (-a) = -a

-3^{\circ \:}x\left(-3\right)=-3\cdot \:14

Multiply the numbers: 3 * 14  = 42

-3^{\circ \:}x\left(-3\right)=-42

Divide both sides by: -3^{\circ \:}\left(-3\right)

\frac{-3^{\circ \:}x\left(-3\right)}{-3^{\circ \:}\left(-3\right)}=\frac{-42}{-3^{\circ \:}\left(-3\right)}

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