)Two similar triangles are shown on the following coordinate grid:

Mathematics

2 answers:

DIA [1.3K]3 years ago

5 0

Answer:

d on edge 2020

Step-by-step explanation:

muminat3 years ago

4 0

Answer:

Reflection across the y-axis and a dilation by a factor of 2.

Step-by-step explanation:

Comparing the points from the pre-image, ABC, to the image, A'B'C', we see that both x- and y-coordinates seem to be doubled from the pre-image to the image. However, the sign of the x-coordinates in the image is different; this means that the x-coordinates are negated. A reflection across the y-axis will negate the x-coordinates.

To double both the x- and y-coordinates, we would then apply a dilation by a factor of 2, as the scale factor gets multiplied by every coordinate.

You might be interested in

Can someone please help me?

Tatiana [17]

It can be solved by factoring or taking the root of both sides.

x= 2 radical 3. and -2 radical 3

7 0

3 years ago

Which of the following equations have complex roots?

goldenfox [79]

9514 1404 393

Answer:

B. 3x² +2 = 0

Step-by-step explanation:

The equation of A has a couple of real roots. We're pretty sure there are complex numbers that will satisfy this equation, but we don't know how to find them. (We suspect a typo, and that the equation is supposed to be 2x² +1 = 7x, which has only real roots.)

The equation of B can be rewritten as ...

x² = -2/3

This will have complex roots.

The discriminants of both equations C and D are positive, so those have only real roots.

2x² -5x -1 ⇒ d = (-5)² -4(2)(-1) = 33

3x² -6x -1 ⇒ d = (-6)² -4(3)(-1) = 48

6 0

3 years ago

3x^2 – 8x + 5 = 5x^2

Ghella [55]

Answer:

Step-by-step explanation:

$3x^2 -8x +5 = 5x^2\\\\5x^2 - 3x^2 + 8x -5 =0\\\\2x^2 +8x -5 = 0\\\\$

a = 2, b = 8, c = -5

$x = \frac{-b \pm \sqrt{b^2 -4ac} }{2a}\\\\x = \frac{-8 \pm \sqrt{8^2 - ( 4 \times 2 \times (-5))} }{2 \times 2}\\\\x = \frac{-8 \pm \sqrt{64 +40} }{4}\\\\x = \frac{-8 + \sqrt{64 +40} }{4} \ , \ x = \frac{-8 - \sqrt{64 +40} }{4}\\\\x = \frac{-8 + \sqrt{104} }{4} \ , \ x = \frac{-8 - \sqrt{104} }{4}\\\\x = \frac{-8 + \sqrt{4 \times 26} }{4} \ , \ x = \frac{-8 - \sqrt{4 \times 26} }{4}\\\\x = \frac{-8 +2 \sqrt{26} }{4} \ , \ x = \frac{-8 - 2\sqrt{26} }{4}\\\\$