So,
123/20 is over 1. The denominator is 20, so what I would do is find out how many <em>whole </em>20's would fit in 123.
The result is that 6 20's would fit in 123 with 3 left over.
The answer is six and three-twentieths (6 3/20). It's already in simplest form.
Answer:
first we will check if the coordinate (5,4) is a solution:
you fill in only the X-coordinate (or the y-coordinate). I fill in 5 for X:
y-4 = 7(5-6)
now we have to work this out like:
y-4 = 7 • -1
y -4 = -7
y = -3 if the coordinate was the solution for the y-coordinate the answer should be 4 but it isn't so coordinate (5,4) isn't a solution.
the same goes for coordinate (6,5):
y-4 = 7 (6-6)
y-4 = 0
y = 4 if the coordinate was the solution for the y-coordinate the answer should be 5 but it isn't so coordinate (6,5) isn't a solution.
so the answer is neither
Property of equality
k=k
10/2=5
5=5
Answer:
Step-by-step explanation:
LOL
So work it out as 0.25x4 =£1
Then 4-1=£3
£3 / 0.30 =10
10+4
So it is 14 days late
