Solve for x:(5 (x - 1/3))/(8) = 5/12
Put each term in x - 1/3 over the common denominator 3: x - 1/3 = (3 x)/3 - 1/3:(5 (3 x)/3 - 1/3)/(8) = 5/12
(3 x)/3 - 1/3 = (3 x - 1)/3:(5 (3 x - 1)/3)/(8) = 5/12
3×8 = 24:(5 (3 x - 1))/24 = 5/12
Multiply both sides of (5 (3 x - 1))/24 = 5/12 by 24/5:(24×5 (3 x - 1))/(5×24) = (24×5)/(5×12)
(24×5)/(5×12) = (24×5)/(5×12):(24×5 (3 x - 1))/(5×24) = (24×5)/(5×12)
(24×5 (3 x - 1))/(5×24) = (5×24)/(5×24)×(3 x - 1) = 3 x - 1:3 x - 1 = (24×5)/(5×12)
(24×5)/(5×12) = 5/5×24/12 = 24/12:3 x - 1 = 24/12
The gcd of 24 and 12 is 12, so 24/12 = (12×2)/(12×1) = 12/12×2 = 2:3 x - 1 = 2
Add 1 to both sides:3 x + (1 - 1) = 1 + 2
1 - 1 = 0:3 x = 2 + 1
2 + 1 = 3:3 x = 3
Divide both sides of 3 x = 3 by 3:(3 x)/3 = 3/3
3/3 = 1:x = 3/3
3/3 = 1:Answer: x = 1
Answer:
9
Step-by-step explanation:
9v = 9(2i + 9j) = 18i + 81j
|9v | = 
= 
= 9
Answer:
A... .................. ....
I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
Remark
Let the pigs = p
Let the chickens = c
Givens
30 heads is another way of saying that there were 30 creatures in the barnyard. In addition each chicken has 2 legs and each pig has 4
c + p = 30
2C + 4P = 100 (2) Divide (2) by 2
C + 2P = 50 (3) Subtract (1) from (3)
<u>C + P = 30</u>
P = 20
So there were 20 pigs.
