5) k/35=3/7 : cross multiply
35*3=7k : simplify
105=7k : divide both sides of the equation by 7
k=15
6) 3/t=18/24 :cross multiply
18t=24*3 :simplify
18t=72 : divide both sides of the equation by 18
t=4
7) 10/8.4 = 5/m :cross multiply
10m=5 * 8.4 :simplify
10m=42 :divide both sides by 10
m=4.2
Answer:
The answer is: There are only 143034 books left after the students return to their classroom
Step-by-step explanation:
Lets call X= Number of books left in the library after the students return to their classroom.
B=Total of Books in the library
S=Total of books taken by the 3rd grade students
If each student of the 3rd grage class take 2 books, the total of book taken by the 3rd grade students would be:
S=2 books per student * Total of students of the 3rdgrade
S=2*21=42
Then we must substract Total of Books in the library minus Total of books taken by the 3rd grade students. That would be:
X=B-S
X=143076-42
X=143034
I hope that this answer will help you
Factor the following:
10 y^2 - 35 y + 30
Factor 5 out of 10 y^2 - 35 y + 30:
5 (2 y^2 - 7 y + 6)
Factor the quadratic 2 y^2 - 7 y + 6.
The coefficient of y^2 is 2 and the constant term is 6.
The product of 2 and 6 is 12.
The factors of 12 which sum to -7 are -3 and -4. So 2 y^2 - 7 y + 6 = 2 y^2 - 4 y - 3 y + 6 = y (2 y - 3) - 2 (2 y - 3):
5 y (2 y - 3) - 2 (2 y - 3)
Factor 2 y - 3 from y (2 y - 3) - 2 (2 y - 3):
Answer: 5 (2 y - 3) (y - 2)
Answer:
4n ≤ 32
/4 /4
n ≤ 8, he could run less than 8 or just 8 laps,
so possible laps:
1, 2, 3, 4, 5, 6, 7, or 8 laps he could've ran.