Answer:
No
Step-by-step explanation:
Neither 4 or 7 is a solution to both equations
Answer:
see in the picture mark brainliest if correct
So have the sequence:

To check if the sequence is geometric, we are going to find its common ratio; to do it we are going to use the formula:

where

is the common ratio

is the current term in the sequence

is the previous term in the sequence
In other words we are going to divide the current term by the previous term a few times, and we will to check if that ratio is the same:
For

and

:


For

and

:


For

and

:


As you can see, we have a common ratio!
We can conclude that our sequence is a geometric sequence and its common ratio is
The number of hours that should be used to study over the weekend =
hrs
The number of hours to study 3 credit course is = 6hrs/ week.
The number of hours studied on monday, wednesday and friday,
= 3/5 + 16/4 +2/3
The lowest common multiple of the denominators 5, 4 and 3 = 60
= 
= 316/60
Reduce to the lowest figure by dividing through with 4,
= 79/15
To detemine the extra hours to be studied over the weekend, subtract 79/15 from 6.
that is, 6/1 - 79/15
The lowest common multiple of denominators 1 and 15 = 15
= 
=
hrs
Learn more here:
brainly.com/question/2166478
Here are the answers to all the questions. I didn't check my answers FYI.