a3=9
assuming the pattern is constant, a3 is right in the middle of a1 and a5 so the answer would be right in the middle of 6 and 12. the average of 6 and 12 is 9, so a3=9.
hope this helps!
Answer:
(a) 7 essays and 29 multiple questions
(b) Your friend is incorrect
Step-by-step explanation:
Represent multiple choice with M and essay with E.
So:
--- Number of questions
--- Points
Solving (a): Number of question of each type.
Make E the subject of formula in 

Substitute 36 - M for E in 


Collect Like Terms


Divide both sides by -4


Substitute 29 for M in 


Solving (b): Can the multiple questions worth 4 points each?
It is not possible.
See explanation.
If multiple question worth 4 points each, then
would be:

Where x represents the number of points for essay questions.
Substitute 7 for E and 29 for M.


Subtract 116 from both sides



Make x the subject

Since the essay question can not have worth negative points.
Then, it is impossible to have the multiple questions worth 4 points
<em>Your friend is incorrect.</em>
Answer:
C. x = 12
Step-by-step explanation:
5(12) + 9 = 69 - 3(12) = 33
18 + 15 = 33
I'll do the first 2 and 6, and I challenge you to do the other three on your own!
For 1, from some guess and check we can figure out that 5*5=25. Since 5 is a prime number, that's it!
For 2, we can figure out that 7*7=49 and 7 is a prime number, so we're good there.
From 6, we can do some guess and check to figure out that 2*24=48, 2*12=24, 2*6=12, and 2*3=6, resulting in 2*2*2*2*3=48 since 2 and 3 are prime numbers. We found out, for example, to find 2*12 due to that if 2*24=48, 2*24 is our current factorization. By finding 2*12=24, we can switch it to 2*2*12