Answer:
The picture is 4.25 inches from the side of the paper
Step-by-step explanation:
- Taylor wants to center a 3.5 inch picture on a piece of paper that is
12 inches wide
- Lets think about that he want to put the picture in the center of the
paper, then divide the length of the paper into two equal parts and the
picture into two equal part
∵ The width of the paper is 12 inches
∵ 12 ÷ 2 = 6 inches
∵ The width of the picture is 3.5 inches
∵ 3.5 ÷ 2 = 1.75
- Now lets subtract from 6 inches (half paper) 1.75 inches (half picture)
to find the distance between the side of the paper and the picture
∵ 6 - 1.75 = 4.25
∴ The distance from the side of the paper to the picture is 4.25 inches
* <em>The picture is 4.25 inches from the side of the paper</em>
* Look to attached figure for more understand
Answer:
90 degrees
Step-by-step explanation:
90 degrees hhgffdfkgdf
Answer:
15 pages
Step-by-step explanation:
This is a fairly simple problem meaning it has very few steps. If it makes 75 pages in 5 minutes you need to find out what the PPM or Page Per Minute is.
To find that you simply divide 75 by 5 and get 15 and in one minute the printer prints 15 pages.
Answer:
8 square units and
square units
Step-by-step explanation:
The area of the triangle ABC is 24 square units.
1. Triangles ABC and FBG are similar with scale factor
then

2. Triangles ABC and DBE are similar with scale factor
then

3. Thus, the area of the quadrilateral DFGE is

and the area of the quadrilateral ADEC is

Answer:0.05168986083
Step-by-step explanation: