Answer:
Step-by-step explanation:
Corresponding angles of both the squares are congruent. (angles of a square measure 90°)
Ratio of the sides of the given squares = ![\frac{\text{Side length of small square}}{\text{Side length of the large square}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BSide%20length%20of%20small%20square%7D%7D%7B%5Ctext%7BSide%20length%20of%20the%20large%20square%7D%7D)
= ![\frac{2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D)
This ratio of side lengths is constant for all corresponding sides.
Therefore, corresponding sides are proportional.
Since, all angles of both the squares are congruent and all the sides are proportional, both the squares will be similar.
Scale factor = ![\frac{\text{Side length of large square}}{\text{Side length of the smaller square}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BSide%20length%20of%20large%20square%7D%7D%7B%5Ctext%7BSide%20length%20of%20the%20smaller%20square%7D%7D)
= ![\frac{5}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D)
= 2.5
This sequence of similarity transformations shows the figures are similar.
42x can be factored into 7(6x), and 49 is square, meaning it can be factored into 7 squared. I believe the factored expression would be 7(6x + 7) because 7 is the greatest common factor between both numbers.
Answer:
the alternate angle is angle 2
Well, notice the composite is really just 4 triangles atop sitting on top of 4 rectangles, and all of them area stacked up at the edges.
so, for the rectangle's sides,
front and back are two 6x3 rectangles
left and right are two 6x3 rectangles
the bottom part is a 6x6 rectangle
now, we don't include the 6x6 rectangle that's touching the triangles, because that's inside area, and is not SURFACE area, so we nevermind that one.
now, the triangles are just four triangles with a base of 6, and a height of 4, in red noted there.
so, just get the area of all those rectangles and the triangles, sum them up and that's the
surface area of the composite,
Answer: 0.4x = 3 is the answer
Step-by-step explanation: