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.
34.215 kilometers you move the decimal to the left three places because Kilo means 1000 and there are 3 zeros in 1000. you move left because Kilometer is bigger than meter.
Answer:
Hey there!
We can write this equation, where x is the hourly rate and y is the fixed fee.
4x+y=270
7x+y=420
4x+y=270
-7x-y=-420
-3x=-150
x=50
4(50)+y=270
200+y=270
y=70.
The fixed fee is G. $70.
Let me know if this helps :)
Answer:
b is the correct answer
Step-by-step explanation:
mark as brainlist
Answer:
C) I and III only
Step-by-step explanation:
Let full pool is denoted by O
Days Hose x takes to fill pool O = a
Pool filled in one day x = O/a
Days Hose y takes to fill pool O = b
Pool filled in one day y = O/b
Days Hose z takes to fill pool O = c
Pool filled in one day z = O/c
It is given that
a>b>c
Days if if x+y+z fill the pool together = d
1 day if x+y+z fill the pool together 
I) d < c
d are days when hose x, y, z are used together where as c are days when only z is used so number of days when three hoses are used together must be less than c when only z hose is used. So d < c
III) 
Using (1)
Similarly
So,
