Answer:
A. X-coordinate
Step-by-step explanation:
I calculated it logically
You can not see the problem
A polygon is a two-dimensional closed object. The polygon that best describes a stop sign is a regular octagon.
<h3>What is a polygon?</h3>
A polygon is a two-dimensional closed object with n number of straight sides that is flat or planar and the value of n is always greater than 2 (n>2).
<h3>What is an octagon?</h3>
An octagon is a polygon that has 8 number of sides.
As we know that a stop sign has 8 sides, therefore, the polygon that is in the shape of the stop sign is an octagon.
An octagon has 8 sides, and as it is mentioned in the problem that sides and angles appear to be congruent, therefore, the polygon must be a regular polygon.
Hence, the polygon that best describes a stop sign is a regular octagon.
Learn more about Polygon:
brainly.com/question/17756657
Answer : 2√3
<u>Given </u><u>:</u><u>-</u>
- A equilateral triangle with side length 4.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The value of x in the given figure.
As we know that in a equilateral triangle , perpendicular bisector , angle bisector and median coincide with each other .
- So the perpendicular drawn in the figure will bisect the given side .
- Therefore the value of each half will be 4/2 = 2 .
Now we may use Pythagoras theorem as ,
→ AB² = BC² + AC²
→ 4² = 2² + x²
→ 16 = 4 + x²
→ x² = 16-4
→ x² = 12
→ x =√12 = √{ 3 * 2²}
→ x = 2√3
<u>Hence </u><u>the</u><u> required</u><u> answer</u><u> is</u><u> </u><u>2</u><u>√</u><u>3</u><u> </u><u>.</u>
I hope this helps.
If Rosie is trying to figure out the amount that she needs to pay for her total yoga classes.
First, we need to establish what we know, and what we do not know.
We know that Rosie knows how many yoga classes she has.
We know that she knows how many yoga classes she bought.
She does not know how much each yoga class costs.
So, that cancels out A, B, and C because she doesn't know how much one yoga class costs.
It can only be D, because D tells us that she knows the total is $30, but not the amount each class costs.