Answer:
Polygon Y's area is one ninth (1/9) of Polygon X's area
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
In this problem
Let
z-----> the scale factor
a-----> Polygon Y's area
b----> Polygon X's area

we have

substitute



therefore
Polygon Y's area is one ninth (1/9) of Polygon X's area
The answer is -9+7 because it INCREASED which means adding ( + ).
Below is the proof why the area of a parallelogram is B times H
<h3>How to form a rectangle from the parallelogram</h3>
The following steps would create a rectangle from the parallelogram
- Step 1: Cut out the triangular part of the parallelogram
- Step 2: Attach the triangular part to the other end of the parallelogram
See attachment for the steps.
When the rectangle has been created, we have the following dimensions:
Length = h
Width = B
The area of a rectangle is:
Area = Length * Width
This gives
Area = h * B
Evaluate
Area = Bh
Hence, the area of a parallelogram is B times H
Read more about parallelograms at:
brainly.com/question/3050890
#SPJ1
Answer:
450
Step-by-step explanation:
513÷114=4.5 then move decimal 2 places to the right
513÷450= 1.14 then move decimal 2 places to the right
<span>4x² - 5x + 1 = 0
a = 4; b = - 5, c = 1
</span>Δ = b² - 4.a.c
Δ = (-5)² - 4.4.1
Δ = 25 - 16
Δ = 9
<span>
x = - b </span>± √Δ / 2.a



A. Two real solutions (x' = -1 and x'' = -1/4)