All categories

3 years ago

Triangle A'B'C' is a rotation of triangle ABC by 60 degrees using Q as the center.

Mathematics

1 answer:

deff fn [24]3 years ago

7 0

9514 1404 393

Answer:

see below

Step-by-step explanation:

When a figure is rotated about a center point, the central angle formed by any point on the figure and the corresponding image point will be the rotation angle. Every image point is the same distance from the center of rotation that the pre-image point was. No lengths or angles are changed: rotation is a "rigid motion", so the rotated figure is congruent with the original.

-Angle AQA' is 60 degrees.

-Triangles ABC and A'B'C' are congruent.

-Angle ABC is congruent to angle A'B'C'. (this is one of the angles of the congruent triangles)

-Segment BC is congruent to segment B'C'

-Segment AQ is congruent to segment A'Q.

You might be interested in

Determine whether these two functions are inverses.

11Alexandr11 [23.1K]

Answer:

No The reactions are not inverses to each other

Step-by-step explanation:

f(x) = 3x + 27

Let f(x) be y

y= 3x+27

subtracting 27 on both sides

3x = y - 27

x= (y-27)/3

= y/3 - 9

inverse function is x/3 -9 not x/3 + 9

Therefore, not an inverse

Hope it helps...

6 0

3 years ago

How to draw the picture

kati45 [8]

Draw four boxes. Put five balloons in one box with the characters in them and write 1/4 under that box. Then draw five balloons in the other boxes, then draw one big box with the total amount of balloons in them.

3 0

3 years ago

Read 2 more answers

4(1-2b)+7b-10

likoan [24]

Answer:

if you're simplifying, it should be = -1b - 6

Step-by-step explanation:

1. 4(1-2b) + 7b -10 solve the parenthesis

2. 4 - 8b + 7b - 10 combine like(same) terms

3. 4 - 1b - 10 same as 2.

4. -1b - 6

5 0

3 years ago

Find the slope of the line that is parallel and perpendicular -7x-2y=4

Rus_ich [418]

$\bf -7x-2y=4\implies -2y=7x+4\implies y=\cfrac{7x+4}{-2}\implies y=\cfrac{7x}{-2}+\cfrac{4}{-2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{7}{2}} x-2\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{7}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{2}{7}}\qquad \stackrel{negative~reciprocal}{\cfrac{2}{7}}}$

now, what's the slope of a line parallel to that one above? well, parallel lines have exactly the same slope.

5 0

3 years ago

Use the cosine sum and difference identities to find the exact value. COS(5pi/12)

nydimaria [60]

5π/12 = (5 · 180°) : 12 = 75°
cos 75° = cos ( 45° + 30° )= cos 45° cos 30° - sin 45° sin 30° =
=√2/2 * √3/2 - √2/2 * 1/2 = $\frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4}= \frac{ \sqrt{6} - \sqrt{2} }{4}$
=(2.4495 - 1.4142): 4 = 0.258825

8 0

2 years ago

Read 2 more answers