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Mathematics

1 answer:

Ulleksa [173]3 years ago

5 0

Following transformations on Triangle ABC will result in the Triangle A'B'C'

a) Reflection the triangle across x-axis
b) Shift towards Right by 2 units
c) Shift upwards by 6 units

In Triangle ABC, the coordinates of the vertices are:
A (1,9)
B (3, 12)
C (4, 4)

In Triangle A'B'C, the coordinates of the vertices are:
A' (3, -3)
B' (5, -6)
C' (6, 2)

First consider point A of Triangle ABC.
Coordinate of A are (1, 9). If we reflect it across x-axis the coordinate of new point will be (1, -9). Moving it 2 units to right will result in the point (3, -9). Moving it 6 units up will result in the point (3,-3) which are the coordinates of point A'.

Coordinates of B are (3,12). Reflecting it across x-axis, we get the new point (3, -12). Moving 2 units towards right, the point is translated to (5, -12). Moving 6 units up we get the point (5, -6), which are the coordinate of B'.

The same way C is translated to C'.

Thus the set of transformations applied on ABC to get A'B'C' are:
a) Reflection the triangle across x-axis
b) Shift towards Right by 2 units
c) Shift upwards by 6 units

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What is the quotient StartFraction 15 p Superscript negative 4 Baseline q Superscript negative 6 Baseline Over negative 20 p Sup

zhannawk [14.2K]

Answer:

$- \frac{3}{4} \times \frac{p^{8} }{q^{3} }$

Step-by-step explanation:

We have to find the quotient of the following division, $\frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}$ .

Now, $\frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}$

= $- \frac{3}{4} p^{[- 4 - (- 12)]} q^{[-6 - (- 3)]}$ {Since all the terms in the expression are in product form, so we can treat them separately}