Answer:
Step-by-step explanation:
Solution:
As given Square L M NO is dilated by a scale factor of two about the center of the square to create square L'M'N'O'.
Original line of Dilation = Along P Q
New Dilated line = P'Q'
As scale factor > 1
1. Image Size > Pre image size
2. The two images will be similar.
3. Length of Dilated Line P' Q' = 2 × Length of PQ
As you can see from the diagram drawn below, Dilated line P'Q' will contain the point P and Q.
All four points P,Q,Q',P' are collinear , lie in the same line.
Option (2) dilated line P'Q' will contain the points P and Q is true.
Answer:
x = 88.2
Step-by-step explanation:
The angle at the top of the triangle = 90° - 10° = 80°
and the left side of the triangle is x ( opposite sides of a rectangle )
Using the tangent ratio in the right triangle
tan80° =
= 
Multiply both sides by x
x × tan80° = 500 ( divide both sides by tan80° )
x =
≈ 88.2
-20%
work:
(24-30):30*100=
(24:30-1)*100=
80-100=-20
Answer:

Step-by-step explanation:
We need to find the distance between two points i.e. (−5,−1) and (−3,−8).
It can be calculated using distance formula as follows :

We have, x₁ = -5, x₂ = -3, y₁ = -1 and y₂ = -8
Put all the values,

So, the distance between the points is equal to 