Answer:
7/10
Step-by-step explanation:
The scale factor that Ivory used to create their scaled copy is;
<u><em>Scale is 1 interval on graph represents 4 units of the rectangle. This is 1:4 ratio.</em></u>
- From the given image, we can see that the y-axis of the rectangle has 3 intervals while the x-axis has 1 intervals
- Now, from the image, we are told that the area of the scaled copy is 48 square units.
What this means is that the area of the triangle she is trying to represent with that graph is 48 square units.
- Now, if one interval of the graph is x-units, then it means;
length; y-axis = 3x units
width; x-axis = x units
- Area of a rectangle is given by;
Area = Length × width
Thus;
3x * x = 48
3x² = 48
x² = 48/3
x² = 16
x = √16
x = 4
<em>Thus, 1 interval on the graph represents 4 units of the rectangle.</em>
Read more at; brainly.com/question/17182684
Answer:
14
Step-by-step explanation:
The diameter of a circle is twice its radius
d=2r
Substitute 7 in for r
d=2*7
d=14
So, the diameter is 14
Hope this helps! :)
Answer:
Step-by-step explanation:
<em>L</em> is 2* <em>W</em> so L=2W
perimeter= 2w+2l
plug in whats given to get
perimeter=2w+4w 4w is because L is *2 of W so that makes L=2w
36=6w
36/6=6
6w/6=w
w=6
6*2=12=L
now we have 6*12=72=Area
We need to determine the radius and diameter of the circle. If the area of the circle is 10 pi in^2, then, according to the formula for the area of a circle,
A = 10 pi in^2 = pi*r^2. Thus, 10 in^2 = r^2, and r = radius of circle = sqrt(10) in.
Thus, the diam. of the circle is 2sqrt(10) in. This diam. has the same length as does the hypotenuse of one of the triangles making up the square.
Thus, [ 2*sqrt(10) ]^2 = x^2 + x^2, where x represents the length of one side of the square. So, 4(10) in^2 = 2x^2. Then:
40 in^2 = 2x^2, or 20 in^2 = x^2, and so the length x of one side of the square is sqrt(20). The area of the square is the square of this result:
Area of the square = x^2 = [ sqrt(20) ]^2 = 20 in^2 (answer). Compare that to the 10 pi sq in area of the circle (31.42 in^2).