Answer:
B. 21.2
Step-by-step explanation:
Perimeter of ∆ABC = AB + BC + AC
A(-4, 1)
B(-2, 3)
C(3, -4)
✔️Distance between A(-4, 1) and B(-2, 3):
AB = 4 units
✔️Distance between B(-2, 3) and C(3, -4):
BC = 8.6 units (nearest tenth)
✔️Distance between A(-4, 1) and C(3, -4):
AC = 8.6 units (nearest tenth)
Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.
Answer would be -4 5/12
Hope that helps.
Die haben sich schon in der Schule gesehen und 5 bis drei z die Frau A bist du noch mal in der Nähe
Answer:
y - 13 = (5/2)(x - 4)
Step-by-step explanation:
Here we know the slope and one point on the line. Use the point-slope formula:
y - k = m(x - h).
Substituting 13 for k, 4 for x and 2.5 for m, we get:
y - 13 = (5/2)(x - 4)\
