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LUCKY_DIMON [66]

3 years ago

12

A^4 x a^-7 Simplify. express using exponents

1 answer:

lara [203]3 years ago

3 0

1/a^3

The guy above me is probably also right

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Triangle QRS has vertices Q(-4,2),R(3,0), and S(4,3).

Dafna1 [17]

The coordinates of the vertices of $\triangle Q'R'S'$ are $Q'(x,y) = (0, -4)$ , $R'(x,y) = (7,-6)$ and $S'(x,y) = (8, -3)$ , respectively.

Vectorially speaking, a point is translated to another location on Cartesian plane by means of this expression:

$A'(x,y) = A(x,y) + T(x,y)$ (1)

Where:

$A(x,y)$ - Original point.
$A'(x,y)$ - Translated point.
$T(x,y)$ - Translation vector.

If we know that $Q(x,y) = (-4,2)$ , $R(x,y) = (3,0)$ , $S(x,y) = (4, 3)$ and $R'(x,y) = (7, -6)$ $T(x,y) = (4, -6)$ , then the coordinates of the vertices of the triangle $Q'R'S'$ are:

$Q'(x,y) = Q(x,y) + T(x,y)$

$Q'(x,y) = (-4,2) + (4,-6)$

$Q'(x,y) = (0, -4)$

$R'(x,y) = R(x,y) + T(x,y)$

$R'(x,y) = (3,0) +(4,-6)$

$R'(x,y) = (7,-6)$

$S'(x,y) = S(x,y) + T(x,y)$

$S'(x,y) = (4, 3) + (4, -6)$

$S'(x,y) = (8, -3)$

The coordinates of the vertices of $\triangle Q'R'S'$ are $Q'(x,y) = (0, -4)$ , $R'(x,y) = (7,-6)$ and $S'(x,y) = (8, -3)$ , respectively.

We kindly invite to see this question on translations: brainly.com/question/17485121

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3 years ago