answersLogoWhite

0

Yes; less than one pixel is 0 pixels which is nothing = no image

Vector images can be of any size but when rendered to screen (or any imaging device which uses pixels) of course the result must be at least 1 pixel for the object to appear. The vector object might very well be there but be invisible due to the (low) resolution of the imaging output.

User Avatar

Wiki User

13y ago

Still curious? Ask our experts.

Chat with our AI personalities

ViviVivi
Your ride-or-die bestie who's seen you through every high and low.
Chat with Vivi
BeauBeau
You're doing better than you think!
Chat with Beau
MaxineMaxine
I respect you enough to keep it real.
Chat with Maxine

Add your answer:

Earn +20 pts
Q: Bitmaps must be greater than one pixel on a side?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Other Math

How do you describe the possible lengths of the third side of a triangle when the other tw sides are given?

The length of the third side must be greater than the difference between the length of the two given sides and it must be less than the sum of the two given sides. These limits can be derived from the fact that any two sides of a triangle must have a combined length greater than the third side.


Is it possible to build a triangle with 9 7 and 16 side lengths?

> No. One side must be greater than the sum of the other two. Actually, each side must be shorter than the sum of the other two in order to build a triangle. In this case, the side of length 16 is not shorter than the sum of the other two sides.


What triangle measures 2m 4m and 7m?

The triangle with side lengths of 2m, 4m, and 7m does not form a valid triangle. In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side according to the Triangle Inequality Theorem. In this case, 2m + 4m is less than 7m, violating the theorem. Therefore, a triangle with these side lengths cannot exist in Euclidean geometry.


What fraction with a denominator of 5 is greater than 3?

This is like solving the folloging inequality (for the unknown numerator): x/5 > 3 Multiplying both side by five gives you: x > 15 So the numerator must be greater than 15.


Is it possible to construct a triangle with the side lengths 30 32 34?

Oh, what a happy little question! It looks like you have a special triangle there with side lengths 30, 32, and 34. Since the sum of the two shorter sides must be greater than the longest side for a triangle to exist, let's check if that's true here. If we add 30 and 32, we get 62, which is indeed greater than 34, so you can paint a beautiful triangle with those side lengths!